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Chapter7<br />
Applications of<br />
Definite Integrals<br />
The art of pottery developed independently<br />
in many ancient civilizations and still exists in<br />
modern times. The desired shape of the side<br />
of a pottery vase can be described by:<br />
y 5.0 2 sin (x/4) (0 x 8p)<br />
where x is the height and y is the radius at<br />
height x (in inches).<br />
A base for the vase is preformed and placed on a<br />
potter’s wheel. How much clay should be added to<br />
the base to form this vase if the inside radius is always<br />
1 inch less than the outside radius? Section 7.3<br />
contains the needed mathematics.<br />
378
Chapter 7 Overview<br />
Section 7.1 Integral as Net Change 379<br />
By this point it should be apparent that finding the limits of Riemann sums is not just an<br />
intellectual exercise; it is a natural way to calculate mathematical or physical quantities<br />
that appear to be irregular when viewed as a whole, but which can be fragmented into regular<br />
pieces. We calculate values for the regular pieces using known formulas, then sum<br />
them to find a value for the irregular whole. This approach to problem solving was around<br />
for thousands of years before calculus came along, but it was tedious work and the more<br />
accurate you wanted to be the more tedious it became.<br />
With calculus it became possible to get exact answers for these problems with almost<br />
no effort, because in the limit these sums became definite integrals and definite integrals<br />
could be evaluated with antiderivatives. With calculus, the challenge became one of fitting<br />
an integrable function to the situation at hand (the “modeling” step) and then finding an<br />
antiderivative for it.<br />
Today we can finesse the antidifferentiation step (occasionally an insurmountable hurdle<br />
for our predecessors) with programs like NINT, but the modeling step is no less crucial.<br />
Ironically, it is the modeling step that is thousands of years old. Before either calculus<br />
or technology can be of assistance, we must still break down the irregular whole into regular<br />
parts and set up a function to be integrated. We have already seen how the process<br />
works with area, volume, and average value, for example. Now we will focus more closely<br />
on the underlying modeling step: how to set up the function to be integrated.<br />
7.1<br />
What you’ll learn about<br />
• Linear Motion Revisited<br />
• General Strategy<br />
• Consumption Over Time<br />
• Net Change from Data<br />
• Work<br />
. . . and why<br />
The integral is a tool that can be<br />
used to calculate net change and<br />
total accumulation.<br />
Integral As Net Change<br />
Linear Motion Revisited<br />
In many applications, the integral is viewed as net change over time. The classic example<br />
of this kind is distance traveled, a problem we discussed in Chapter 5.<br />
EXAMPLE 1<br />
Interpreting a Velocity Function<br />
Figure 7.1 shows the velocity<br />
d s<br />
v(t) t<br />
dt<br />
2 c<br />
<br />
t <br />
81 2 m s ec<br />
of a particle moving along a horizontal s-axis for 0 t 5. Describe the motion.<br />
SOLUTION<br />
Solve Graphically The graph of v (Figure 7.1) starts with v0 8, which we interpret<br />
as saying that the particle has an initial velocity of 8 cmsec to the left. It slows to a<br />
halt at about t 1.25 sec, after which it moves to the right v 0 with increasing speed,<br />
reaching a velocity of v524.8 cmsec at the end.<br />
Now try Exercise 1(a).<br />
[0, 5] by [–10, 30]<br />
Figure 7.1 The velocity function in<br />
Example 1.<br />
EXAMPLE 2<br />
Finding Position from Displacement<br />
Suppose the initial position of the particle in Example 1 is s0 9. What is the particle’s<br />
position at (a) t 1 sec? (b) t 5 sec?<br />
SOLUTION<br />
Solve Analytically<br />
(a) The position at t 1 is the initial position s0 plus the displacement (the<br />
amount, Δs, that the position changed from t 0 to t 1). When velocity is<br />
continued
380 Chapter 7 Applications of Definite Integrals<br />
Reminder from Section 3.4<br />
A change in position is a displacement.<br />
If s(t) is a body’s position at time t, the<br />
displacement over the time interval<br />
from t to t Δt is s(t Δt) s(t). The<br />
displacement may be positive, negative,<br />
or zero, depending on the motion.<br />
constant during a motion, we can find the displacement (change in position) with the<br />
formula<br />
Displacement rate of change time.<br />
But in our case the velocity varies, so we resort instead to partitioning the time interval<br />
0, 1 into subintervals of length Δt so short that the velocity is effectively constant on<br />
each subinterval. If t k is any time in the kth subinterval, the particle’s velocity throughout<br />
that interval will be close to vt k . The change in the particle’s position during the brief<br />
time this constant velocity applies is<br />
vt k Δt. rate of change time<br />
If vt k is negative, the displacement is negative and the particle will move left. If vt k <br />
is positive, the particle will move right. The sum<br />
vt k Δt<br />
of all these small position changes approximates the displacement for the time interval<br />
0, 1.<br />
The sum vt k Δt is a Riemann sum for the continuous function vt over 0, 1. As<br />
the norms of the partitions go to zero, the approximations improve and the sums converge<br />
to the integral of v over 0, 1, giving<br />
Displacement 1<br />
vt dt<br />
0<br />
( t2<br />
0<br />
1<br />
) dt<br />
t <br />
8<br />
1 2<br />
]<br />
<br />
[ t 3<br />
8<br />
<br />
3 t 1<br />
1 3 8 2 8 1 1<br />
.<br />
3<br />
1<br />
0<br />
During the first second of motion, the particle moves 113 cm to the left. It starts at<br />
s0 9, so its position at t 1 is<br />
New position initial position displacement 9 1 1<br />
1 6 .<br />
3 3<br />
(b) If we model the displacement from t 0 to t 5 in the same way, we arrive at<br />
Displacement 5<br />
5<br />
vt dt <br />
[ t 3<br />
8<br />
<br />
3<br />
0<br />
t <br />
]<br />
35.<br />
1<br />
0<br />
The motion has the net effect of displacing the particle 35 cm to the right of its starting<br />
point. The particle’s final position is<br />
Final position initial position displacement<br />
s0 35 9 35 44.<br />
]<br />
Support Graphically The position of the particle at any time t is given by<br />
st t<br />
8<br />
[ u2 <br />
0 u 1 2 du 9,<br />
because st vt and s0 9. Figure 7.2 shows the graph of st given by the<br />
parametrization<br />
xt NINT vu, u, 0,t 9, yt t, 0 t 5.<br />
continued
Section 7.1 Integral as Net Change 381<br />
(a) Figure 7.2a supports that the position of the particle at t 1 is 163.<br />
(b) Figure 7.2b shows the position of the particle is 44 at t 5. Therefore, the displacement<br />
is 44 9 35.<br />
Now try Exercise 1(b).<br />
T = 1<br />
X = 5.3333333 Y = 1<br />
[–10, 50] by [–2, 6]<br />
(a)<br />
T = 5<br />
X = 44 Y = 5<br />
[–10, 50] by [–2, 6]<br />
(b)<br />
Figure 7.2 Using TRACE and the<br />
parametrization in Example 2 you can<br />
“see” the left and right motion of the<br />
particle.<br />
The reason for our method in Example 2 was to illustrate the modeling step that will<br />
be used throughout this chapter. We can also solve Example 2 using the techniques of<br />
Chapter 6 as shown in Exploration 1.<br />
EXPLORATION 1<br />
Revisiting Example 2<br />
The velocity of a particle moving along a horizontal s-axis for 0 t 5 is<br />
d s<br />
t<br />
dt<br />
2 .<br />
t <br />
81 2<br />
1. Use the indefinite integral of dsdt to find the solution of the initial value<br />
problem<br />
d s<br />
t<br />
dt<br />
2 , t <br />
81 2 s0 9.<br />
2. Determine the position of the particle at t 1. Compare your answer with the<br />
answer to Example 2a.<br />
3. Determine the position of the particle at t 5. Compare your answer with the<br />
answer to Example 2b.<br />
We know now that the particle in Example 1 was at s0 9 at the beginning of the<br />
motion and at s5 44 at the end. But it did not travel from 9 to 44 directly—it began its<br />
trip by moving to the left (Figure 7.2). How much distance did the particle actually travel?<br />
We find out in Example 3.<br />
EXAMPLE 3<br />
Calculating Total Distance Traveled<br />
Find the total distance traveled by the particle in Example 1.<br />
SOLUTION<br />
Solve Analytically We partition the time interval as in Example 2 but record every<br />
position shift as positive by taking absolute values. The Riemann sum approximating<br />
total distance traveled is<br />
vt k Δt,<br />
and we are led to the integral<br />
Total distance traveled 5<br />
vt dt 5<br />
8<br />
0<br />
0 t2 t 1 2 dt.<br />
Evaluate Numerically We have<br />
NINT ( t 2 t <br />
81 2 , t, 0,5) 42.59. Now try Exercise 1(c).
382 Chapter 7 Applications of Definite Integrals<br />
What we learn from Examples 2 and 3 is this: Integrating velocity gives displacement<br />
(net area between the velocity curve and the time axis). Integrating the absolute value of<br />
velocity gives total distance traveled (total area between the velocity curve and the time<br />
axis).<br />
General Strategy<br />
The idea of fragmenting net effects into finite sums of easily estimated small changes is<br />
not new. We used it in Section 5.1 to estimate cardiac output, volume, and air pollution.<br />
What is new is that we can now identify many of these sums as Riemann sums and express<br />
their limits as integrals. The advantages of doing so are twofold. First, we can evaluate one<br />
of these integrals to get an accurate result in less time than it takes to crank out even the<br />
crudest estimate from a finite sum. Second, the integral itself becomes a formula that enables<br />
us to solve similar problems without having to repeat the modeling step.<br />
The strategy that we began in Section 5.1 and have continued here is the following:<br />
Strategy for Modeling with Integrals<br />
1. Approximate what you want to find as a Riemann sum of values of a continuous<br />
function multiplied by interval lengths. If f x is the function and a, b the interval,<br />
and you partition the interval into subintervals of length Δx, the approximating<br />
sums will have the form f c k Δx with c k a point in the kth subinterval.<br />
2. Write a definite integral, here b f x dx, to express the limit of these sums as<br />
a<br />
the norms of the partitions go to zero.<br />
3. Evaluate the integral numerically or with an antiderivative.<br />
EXAMPLE 4<br />
Modeling the Effects of Acceleration<br />
A car moving with initial velocity of 5 mph accelerates at the rate of at 2.4t<br />
mph per second for 8 seconds.<br />
(a) How fast is the car going when the 8 seconds are up?<br />
(b) How far did the car travel during those 8 seconds?<br />
SOLUTION<br />
(a) We first model the effect of the acceleration on the car’s velocity.<br />
Step 1: Approximate the net change in velocity as a Riemann sum. When acceleration<br />
is constant,<br />
velocity change acceleration time applied. rate of change time<br />
To apply this formula, we partition 0, 8 into short subintervals of length Δt. On<br />
each subinterval the acceleration is nearly constant, so if t k is any point in the kth<br />
subinterval, the change in velocity imparted by the acceleration in the subinterval is<br />
approximately<br />
at k Δt mph. m ph<br />
sec<br />
sec<br />
The net change in velocity for 0 t 8 is approximately<br />
at k Δt mph.<br />
Step 2: Write a definite integral. The limit of these sums as the norms of the partitions<br />
go to zero is<br />
8<br />
at dt.<br />
0<br />
continued
Section 7.1 Integral as Net Change 383<br />
Step 3: Evaluate the integral. Using an antiderivative, we have<br />
Net velocity change 2]<br />
8<br />
8<br />
2.4tdt 1.2t 76.8 mph.<br />
0<br />
0<br />
So, how fast is the car going when the 8 seconds are up? Its initial velocity is 5 mph and<br />
the acceleration adds another 76.8 mph for a total of 81.8 mph.<br />
(b) There is nothing special about the upper limit 8 in the preceding calculation. Applying<br />
the acceleration for any length of time t adds<br />
t<br />
2.4u du mph u is just a dummy variable here.<br />
0<br />
(b) to the car’s velocity, giving<br />
vt 5 t<br />
The distance traveled from t 0 to t 8 sec is<br />
8<br />
0<br />
0<br />
2.4u du 5 1.2t 2 mph.<br />
vt dt 8<br />
5 1.2t 2 dt Extension of Example 3<br />
0<br />
8<br />
<br />
[<br />
5t 0.4t 3]<br />
0<br />
244.8 mph seconds.<br />
Miles-per-hour second is not a distance unit that we normally work with! To convert to<br />
miles we multiply by hourssecond 13600, obtaining<br />
1<br />
244.8 0.068 mile.<br />
36 00<br />
m i h<br />
sec mi<br />
h se c<br />
The car traveled 0.068 mi during the 8 seconds of acceleration. Now try Exercise 9.<br />
Consumption Over Time<br />
The integral is a natural tool to calculate net change and total accumulation of more quantities<br />
than just distance and velocity. Integrals can be used to calculate growth, decay, and,<br />
as in the next example, consumption. Whenever you want to find the cumulative effect of a<br />
varying rate of change, integrate it.<br />
EXAMPLE 5<br />
Potato Consumption<br />
From 1970 to 1980, the rate of potato consumption in a particular country was Ct <br />
2.2 1.1 t millions of bushels per year, with t being years since the beginning of 1970.<br />
How many bushels were consumed from the beginning of 1972 to the end of 1973?<br />
SOLUTION<br />
We seek the cumulative effect of the consumption rate for 2 t 4.<br />
Step 1: Riemann sum. We partition 2, 4 into subintervals of length Δt and let t k be a time<br />
in the kth subinterval. The amount consumed during this interval is approximately<br />
Ct k Δt million bushels.<br />
The consumption for 2 t 4 is approximately<br />
Ct k Δt million bushels.<br />
continued
384 Chapter 7 Applications of Definite Integrals<br />
Step 2: Definite integral. The amount consumed from t 2 to t 4 is the limit of<br />
these sums as the norms of the partitions go to zero.<br />
Ct dt 4<br />
2.2 1.1 t dt million bushels<br />
4<br />
2<br />
2<br />
Step 3: Evaluate. Evaluating numerically, we obtain<br />
NINT 2.2 1.1 t , t, 2,47.066 million bushels.<br />
Now try Exercise 21.<br />
Table 7.1<br />
Joules<br />
Time (min)<br />
Pumping Rates<br />
Rate (galmin)<br />
0 58<br />
5 60<br />
10 65<br />
15 64<br />
20 58<br />
25 57<br />
30 55<br />
35 55<br />
40 59<br />
45 60<br />
50 60<br />
55 63<br />
60 63<br />
The joule, abbreviated J and pronounced<br />
“jewel,” is named after the<br />
English physicist James Prescott Joule<br />
(1818–1889). The defining equation is<br />
1 joule (1 newton)(1 meter).<br />
In symbols, 1 J 1 N • m.<br />
It takes a force of about 1 N to lift an<br />
apple from a table. If you lift it 1 m you<br />
have done about 1 J of work on the<br />
apple. If you eat the apple, you will have<br />
consumed about 80 food calories, the<br />
heat equivalent of nearly 335,000<br />
joules. If this energy were directly useful<br />
for mechanical work (it’s not), it would<br />
enable you to lift 335,000 more apples<br />
up 1 m.<br />
Net Change from Data<br />
Many real applications begin with data, not a fully modeled function. In the next example,<br />
we are given data on the rate at which a pump operates in consecutive 5-minute intervals<br />
and asked to find the total amount pumped.<br />
EXAMPLE 6<br />
Finding Gallons Pumped from Rate Data<br />
A pump connected to a generator operates at a varying rate, depending on how much<br />
power is being drawn from the generator to operate other machinery. The rate (gallons<br />
per minute) at which the pump operates is recorded at 5-minute intervals for one hour<br />
as shown in Table 7.1. How many gallons were pumped during that hour?<br />
SOLUTION<br />
Let Rt,0 t 60, be the pumping rate as a continuous function of time for the hour.<br />
We can partition the hour into short subintervals of length Δt on which the rate is nearly<br />
constant and form the sum Rt k Δt as an approximation to the amount pumped during<br />
the hour. This reveals the integral formula for the number of gallons pumped to be<br />
Gallons pumped 60<br />
Rt dt.<br />
0<br />
We have no formula for R in this instance, but the 13 equally spaced values in Table 7.1<br />
enable us to estimate the integral with the Trapezoidal Rule:<br />
60<br />
Rt dt 2 [ 58 260 265 … 263 63]<br />
60<br />
0<br />
• 12<br />
3582.5.<br />
The total amount pumped during the hour is about 3580 gal. Now try Exercise 27.<br />
Work<br />
In everyday life, work means an activity that requires muscular or mental effort. In science,<br />
the term refers specifically to a force acting on a body and the body’s subsequent displacement.<br />
When a body moves a distance d along a straight line as a result of the action of a<br />
force of constant magnitude F in the direction of motion, the work done by the force is<br />
W Fd.<br />
The equation W Fd is the constant-force formula for work.<br />
The units of work are force distance. In the metric system, the unit is the newtonmeter,<br />
which, for historical reasons, is called a joule (see margin note). In the U.S. customary<br />
system, the most common unit of work is the foot-pound.
Section 7.1 Integral as Net Change 385<br />
Hooke’s Law for springs says that the force it takes to stretch or compress a spring x<br />
units from its natural (unstressed) length is a constant times x. In symbols,<br />
F kx,<br />
where k, measured in force units per unit length, is a characteristic of the spring called the<br />
force constant.<br />
EXAMPLE 7 A Bit of Work<br />
It takes a force of 10 N to stretch a spring 2 m beyond its natural length. How much work<br />
is done in stretching the spring 4 m from its natural length?<br />
SOLUTION<br />
We let Fx represent the force in newtons required to stretch the spring x meters from<br />
its natural length. By Hooke’s Law, Fx kx for some constant k. We are told that<br />
The force required to stretch the<br />
F2 10 k • 2,<br />
spring 2 m is 10 newtons.<br />
so k 5 Nm and Fx 5x for this particular spring.<br />
We construct an integral for the work done in applying F over the interval from x 0<br />
to x 4.<br />
Step 1: Riemann sum. We partition the interval into subintervals on each of which F is<br />
so nearly constant that we can apply the constant-force formula for work. If x k<br />
is any point in the kth subinterval, the value of F throughout the interval is<br />
approximately Fx k 5x k . The work done by F across the interval is<br />
approximately 5x k Δx, where Δx is the length of the interval. The sum<br />
Numerically, work is the area under the<br />
force graph.<br />
Fx k Δx 5x k Δx<br />
approximates the work done by F from x 0 to x 4.<br />
Steps 2 and 3: Integrate. The limit of these sums as the norms of the partitions go to<br />
zero is<br />
4<br />
Fx dx 4<br />
5xdx 5 x 4<br />
2<br />
40 N • m.<br />
2<br />
0<br />
0<br />
]<br />
0<br />
Now try Exercise 29.<br />
We will revisit work in Section 7.5.<br />
Quick Review 7.1 (For help, go to Section 1.2.)<br />
In Exercises 1–10, find all values of x (if any) at which the function<br />
changes sign on the given interval. Sketch a number line graph of the<br />
interval, and indicate the sign of the function on each subinterval.<br />
Example: f x x 2 1 on 2, 3<br />
f (x)<br />
+ – +<br />
–2 –1 1 3<br />
Changes sign at x 1.<br />
1. sin 2x on 3, 2 Changes sign at p 2 ,0, p 2 <br />
2. x 2 3x 2 on 2, 4 Changes sign at 1, 2<br />
x<br />
3. x 2 2x 3 on 4, 2 Always positive<br />
4. 2x 3 3x 2 1 on 2, 2 Changes sign at 1 2 <br />
5. x cos 2x on 0, 4 Changes sign at p 4 , 3 p 5p , <br />
4 4<br />
6. xe x on 0, ∞ Always positive<br />
x<br />
7. x 2 on<br />
1<br />
5, 30 Changes sign at 0<br />
8. x 2<br />
2<br />
x2<br />
on<br />
4<br />
3, 3 Changes sign at 2, 2, 2, 2<br />
9. sec 1 1 sin x 2 on ∞, ∞<br />
1<br />
10. sin 1x on 0.1, 0.2 Changes sign at 3 p , 1<br />
2p <br />
9. Changes sign at 0.9633 kp, 2.1783 kp, where k is an integer
386 Chapter 7 Applications of Definite Integrals<br />
Section 7.1 Exercises<br />
In Exercises 1–8, the function v(t) is the velocity in msec of a<br />
particle moving along the x-axis. Use analytic methods to do each of<br />
the following:<br />
(a) Determine when the particle is moving to the right, to the<br />
left, and stopped.<br />
(b) Find the particle’s displacement for the given time interval. If<br />
s(0) 3, what is the particle’s final position?<br />
(c) Find the total distance traveled by the particle.<br />
1. vt 5 cos t, 0 t 2p See page 389.<br />
2. vt 6 sin 3t, 0 t p2 See page 389.<br />
3. vt 49 9.8t, 0 t 10 See page 389.<br />
4. vt 6t 2 18t 12, 0 t 2 See page 389.<br />
5. vt 5 sin 2 t cos t, 0 t 2p See page 389.<br />
6. vt 4 t, 0 t 4 See page 389.<br />
7. vt e sin t cos t, 0 t 2p See page 389.<br />
t<br />
8. vt , 0 t 3 See page 389.<br />
1 <br />
t 2<br />
9. An automobile accelerates from rest at 1 3t mphsec for<br />
9 seconds.<br />
(a) What is its velocity after 9 seconds? 63 mph<br />
(b) How far does it travel in those 9 seconds?<br />
344.52 feet<br />
10. A particle travels with velocity<br />
vt t 2 sin t msec<br />
for 0 t 4 sec.<br />
(a) What is the particle’s displacement? 1.44952 meters<br />
(b) What is the total distance traveled?<br />
1.91411 meters<br />
11. Projectile Recall that the acceleration due to Earth’s gravity is<br />
32 ftsec 2 . From ground level, a projectile is fired straight<br />
upward with velocity 90 feet per second.<br />
(a) What is its velocity after 3 seconds? 6 ft/sec<br />
(b) When does it hit the ground?<br />
5.625 sec<br />
(c) When it hits the ground, what is the net distance it has<br />
traveled? 0<br />
(d) When it hits the ground, what is the total distance it has<br />
traveled? 253.125 feet<br />
In Exercises 12–16, a particle moves along the x-axis (units in cm).<br />
Its initial position at t 0 sec is x0 15. The figure shows the<br />
graph of the particle’s velocity vt. The numbers are the areas of<br />
the enclosed regions.<br />
4<br />
5<br />
a b c<br />
24<br />
12. What is the particle’s displacement between t 0 and t c?<br />
23 cm<br />
13. What is the total distance traveled by the particle in the same<br />
time period? 33 cm<br />
a:11 b:16 c: 8<br />
14. Give the positions of the particle at times a, b, and c.<br />
15. Approximately where does the particle achieve its greatest<br />
positive acceleration on the interval 0, b? t a<br />
16. Approximately where does the particle achieve its greatest<br />
positive acceleration on the interval 0, c? t c<br />
In Exercises 17–20, the graph of the velocity of a particle moving on<br />
the x-axis is given. The particle starts at x 2 when t 0.<br />
(a) Find where the particle is at the end of the trip.<br />
(b) Find the total distance traveled by the particle.<br />
17. v (m/sec)<br />
(a) 6 (b) 4 meters<br />
18.<br />
19.<br />
20.<br />
2<br />
0<br />
1<br />
0<br />
–1<br />
2<br />
3<br />
v (m/sec)<br />
0 1<br />
–3<br />
v (m/sec)<br />
0 1<br />
–2<br />
v (m/sec)<br />
1 2 3 4<br />
1 2 3 4<br />
2 3 4 5 6 7<br />
2 3 4 5 6 7<br />
t (sec)<br />
t (sec)<br />
8 9 10<br />
t (sec)<br />
(a) 2<br />
(a) 5<br />
t (sec)<br />
(b) 4 meters<br />
(b) 7 meters<br />
(a) 2.5<br />
(b) 19.5 meters<br />
21. U.S. Oil Consumption The rate of consumption of oil in the<br />
United States during the 1980s (in billions of barrels per year)<br />
is modeled by the function C 27.08 • e t25 , where t is the<br />
number of years after January 1, 1980. Find the total consumption<br />
of oil in the United States from January 1, 1980 to January 1,<br />
1990. 332.965 billion barrels<br />
22. Home Electricity Use The rate at which your home consumes<br />
electricity is measured in kilowatts. If your home consumes<br />
electricity at the rate of 1 kilowatt for 1 hour, you will be charged
26. (b) The answer in (a) corresponds to the area of midpoint rectangles. Part<br />
of each rectangle is above the curve and part is below.<br />
for 1 “kilowatt-hour” of electricity. Suppose that the average<br />
consumption rate for a certain home is modeled by the function<br />
Ct 3.9 2.4 sin pt12, where Ct is measured in<br />
kilowatts and t is the number of hours past midnight. Find the<br />
average daily consumption for this home, measured in kilowatthours.<br />
93.6 kilowatt-hours<br />
23. Population Density Population density measures the number<br />
of people per square mile inhabiting a given living area.<br />
Washerton’s population density, which decreases as you move<br />
away from the city center, can be approximated by the function<br />
10,0002 r at a distance r miles from the city center.<br />
(a) If the population density approaches zero at the edge of the<br />
city, what is the city’s radius? 2 miles<br />
(b) A thin ring around the center of the city has thickness Δr and<br />
radius r. If you straighten it out, it suggests a rectangular strip.<br />
Approximately what is its area? 2prr<br />
(c) Writing to Learn Explain why the population of the ring<br />
in part (b) is approximately<br />
10,0002 r2pr Δr.<br />
Population Population density Area<br />
(d) Estimate the total population of Washerton by setting up<br />
and evaluating a definite integral. 83,776<br />
24. Oil Flow Oil flows through a cylindrical pipe of radius<br />
3 inches, but friction from the pipe slows the flow toward the<br />
outer edge. The speed at which the oil flows at a distance r<br />
inches from the center is 810 r 2 inches per second.<br />
(a) In a plane cross section of the pipe, a thin ring with thickness<br />
Δr at a distance r inches from the center approximates a<br />
rectangular strip when you straighten it out. What is the area<br />
of the strip (and hence the approximate area of the ring)? 2prr<br />
(b) Explain why we know that oil passes through this ring at<br />
approximately 810 r 2 2pr Δr cubic inches per second.<br />
(c) Set up and evaluate a definite integral that will give the rate<br />
(in cubic inches per second) at which oil is flowing through the<br />
pipe. 396p in 3 /sec or 1244.07 in 3 /sec<br />
25. Group Activity Bagel Sales From 1995 to 2005, the<br />
Konigsberg Bakery noticed a consistent increase in annual sales<br />
of its bagels. The annual sales (in thousands of bagels) are<br />
shown below.<br />
Sales<br />
Year<br />
(thousands)<br />
1995 5<br />
1996 8.9<br />
1997 16<br />
1998 26.3<br />
1999 39.8<br />
2000 56.5<br />
2001 76.4<br />
2002 99.5<br />
2003 125.8<br />
2004 155.3<br />
2005 188<br />
24. (b) 8(10 r 2 ) in/sec (2pr)r in 2 flow in in 3 /sec<br />
Section 7.1 Integral as Net Change 387<br />
(a) What was the total number of bagels sold over the 11-year<br />
period? (This is not a calculus question!) 797.5 thousand<br />
(b) Use quadratic regression to model the annual bagel sales (in<br />
thousands) as a function Bx, where x is the number of years<br />
after 1995. B(x) 1.6x 2 2.3x 5.0<br />
(c) Integrate Bx over the interval 0, 11 to find total bagel<br />
sales for the 11-year period. 904.02<br />
(d) Explain graphically why the answer in part (a) is smaller<br />
than the answer in part (c). See page 389.<br />
26. Group Activity (Continuation of Exercise 25)<br />
(a) Integrate Bx over the interval 0.5, 10.5 to find total<br />
bagel sales for the 11-year period. 798.97 thousand<br />
(b) Explain graphically why the answer in part (a) is better than<br />
the answer in 25(c).<br />
27. Filling Milk Cartons A machine fills milk cartons with milk<br />
at an approximately constant rate, but backups along the assembly<br />
line cause some variation. The rates (in cases per hour) are<br />
recorded at hourly intervals during a 10-hour period, from<br />
8:00 A.M. to 6:00 P.M.<br />
Rate<br />
Time<br />
(casesh)<br />
8 120<br />
9 110<br />
10 115<br />
11 115<br />
12 119<br />
1 120<br />
2 120<br />
3 115<br />
4 112<br />
5 110<br />
6 121<br />
Use the Trapezoidal Rule with n 10 to determine approximately<br />
how many cases of milk were filled by the machine over<br />
the 10-hour period. 1156.5<br />
28. Writing to Learn As a school project, Anna accompanies her<br />
mother on a trip to the grocery store and keeps a log of the car’s<br />
speed at 10-second intervals. Explain how she can use the data<br />
to estimate the distance from her home to the store. What is the<br />
connection between this process and the definite integral?<br />
See page 389.<br />
29. Hooke’s Law A certain spring requires a force of 6 N to<br />
stretch it 3 cm beyond its natural length.<br />
(a) What force would be required to stretch the string 9 cm<br />
beyond its natural length? 18 N<br />
(b) What would be the work done in stretching the string 9 cm<br />
beyond its natural length? 81 N cm<br />
30. Hooke’s Law Hooke’s Law also applies to compressing springs;<br />
that is, it requires a force of kx to compress a spring a distance x<br />
from its natural length. Suppose a 10,000-lb force compressed a<br />
spring from its natural length of 12 inches to a length of 11 inches.<br />
How much work was done in compressing the spring<br />
(a) the first half-inch? (b) the second half-inch?<br />
1250 inch-pounds 3750 inch-pounds
388 Chapter 7 Applications of Definite Integrals<br />
Standardized Test Questions<br />
You may use a graphing calculator to solve the following<br />
problems.<br />
31. True or False The figure below shows the velocity for a particle<br />
moving along the x-axis. The displacement for this particle is<br />
negative. Justify your answer. False. The displacement is the<br />
integral of the velocity from t 0 to t 5 and is positive.<br />
2<br />
1<br />
32. True or False If the velocity of a particle moving along the<br />
x-axis is always positive, then the displacement is equal to the<br />
total distance traveled. Justify your answer.<br />
33. Multiple Choice The graph below shows the rate at which<br />
water is pumped from a storage tank. Approximate the total<br />
gallons of water pumped from the tank in 24 hours. C<br />
(A) 600 (B) 2400 (C) 3600 (D) 4200 (E) 4800<br />
250<br />
200<br />
150<br />
100<br />
v (m/sec)<br />
0 1<br />
–1<br />
–2<br />
50<br />
r (gal/hr)<br />
2 3 4 5<br />
0 6 12 18 24<br />
t (sec)<br />
t (hr)<br />
34. Multiple Choice The data for the acceleration a(t) of a car<br />
from 0 to 15 seconds are given in the table below. If the velocity<br />
at t 0 is 5 ft/sec, which of the following gives the approximate<br />
velocity at t 15 using the Trapezoidal Rule? D<br />
(A) 47 ft/sec (B) 52 ft/sec (C) 120 ft/sec<br />
(D) 125 ft/sec (E) 141 ft/sec<br />
t (sec) 0 3 6 9 12 15<br />
a(t) (ftsec 2 ) 4 8 6 9 10 10<br />
35. Multiple Choice The rate at which customers arrive at a<br />
counter to be served is modeled by the function F defined by<br />
F(t) 12 6 cos <br />
p<br />
t for 0 t 60, where F(t) is measured in<br />
customers per minute and t is measured in minutes. To the nearest<br />
whole number, how many customers arrive at the counter<br />
over the 60-minute period? B<br />
(A) 720 (B) 725 (C) 732 (D) 744 (E) 756<br />
6<br />
36. Multiple Choice Pollution is being removed from a lake at a<br />
rate modeled by the function y 20e 0.5t tons/yr, where t is the<br />
number of years since 1995. Estimate the amount of pollution removed<br />
from the lake between 1995 and 2005. Round your answer<br />
to the nearest ton. A<br />
(A) 40 (B) 47 (C) 56 (D) 61 (E) 71<br />
Extending the Ideas<br />
37. Inflation Although the economy is continuously changing,<br />
we analyze it with discrete measurements. The following table<br />
records the annual inflation rate as measured each month for<br />
13 consecutive months. Use the Trapezoidal Rule with n 12<br />
to find the overall inflation rate for the year. 0.04875<br />
Month<br />
Annual Rate<br />
January 0.04<br />
February 0.04<br />
March 0.05<br />
April 0.06<br />
May 0.05<br />
June 0.04<br />
July 0.04<br />
August 0.05<br />
September 0.04<br />
October 0.06<br />
November 0.06<br />
December 0.05<br />
January 0.05<br />
38. Inflation Rate The table below shows the monthly inflation<br />
rate (in thousandths) for energy prices for thirteen consecutive<br />
months. Use the Trapezoidal Rule with n 12 to approximate<br />
the annual inflation rate for the 12-month period running<br />
from the middle of the first month to the middle of the last<br />
month. 40 thousandths or 0.040<br />
Monthly Rate<br />
Month (in thousandths)<br />
January 3.6<br />
February 4.0<br />
March 3.1<br />
April 2.8<br />
May 2.8<br />
June 3.2<br />
July 3.3<br />
August 3.1<br />
September 3.2<br />
October 3.4<br />
November 3.4<br />
December 3.9<br />
January 4.0<br />
32. True. Since the velocity is positive, the integral of the velocity is equal to<br />
the integral of its absolute value, which is the total distance traveled.
Section 7.1 Integral as Net Change 389<br />
39. Center of Mass Suppose we have a finite collection of masses<br />
in the coordinate plane, the mass m k located at the point x k , y k <br />
as shown in the figure.<br />
O<br />
Each mass m k has moment m k y k with respect to the x-axis and<br />
moment m k x k about the y-axis. The moments of the<br />
entire system with respect to the two axes are<br />
Moment about x-axis: M x <br />
m k y k ,<br />
Moment about y-axis: M y <br />
m k x k .<br />
The center of mass is xJ, yJ where<br />
xJ M y m<br />
and<br />
M<br />
yJ M k x x m k y<br />
k<br />
k<br />
.<br />
m k<br />
M m k<br />
y<br />
y k x k m k(xk<br />
, y k )<br />
y k<br />
x k<br />
x<br />
Suppose we have a thin, flat plate occupying a region in<br />
the plane.<br />
(a) Imagine the region cut into thin strips parallel to the<br />
y-axis. Show that<br />
xJ x dm<br />
,<br />
dm<br />
where dm d dA, d density (mass per unit area), and<br />
A area of the region.<br />
(b) Imagine the region cut into thin strips parallel to the<br />
x-axis. Show that<br />
yJ y dm<br />
,<br />
dm<br />
where dm d dA, d density, and A area of the region.<br />
In Exercises 40 and 41, use Exercise 39 to find the center of mass<br />
of the region with given density.<br />
40. the region bounded by the parabola y x 2 and the line y 4<br />
with constant density d x 0, y 12/5<br />
41. the region bounded by the lines y x, y x, x 2 with<br />
constant density d x 4/3, y 0<br />
1. (a) Right: 0 t p/2, 3p/2 t 2p<br />
Left: p/2 t 3p/2<br />
Stopped: t p/2, 3p/2<br />
(b) 0; 3 (c) 20<br />
2. (a) Right: 0 t p/3<br />
Left: p/3 t p/2<br />
Stopped: t 0, p/3<br />
(b) 2; 5 (c) 6<br />
3. (a) Right: 0 t 5<br />
Left: 5 t 10<br />
Stopped: t 5<br />
(b) 0; 3 (c) 245<br />
4. (a) Right: 0 t 1<br />
Left: 1 t 2<br />
Stopped: t 1, 2<br />
(b) 4; 7 (c) 6<br />
5. (a) Right: 0 t p/2, 3p/2 t 2p<br />
Left: p/2 t p, p t 3p/2<br />
Stopped: t 0, p/2, p, 3p/2, 2p<br />
(b) 0 ; 3 (c) 20/3<br />
6. (a) Right: 0 t 4<br />
Left: never<br />
Stopped: t 4<br />
(b) 16/3; 25/3 (c) 16/3<br />
25. (d) The answer in (a) corresponds to the area of left hand rectangles.<br />
These rectangles lie under the curve B(x). The answer in (c) corresponds<br />
to the area under the curve. This area is greater than the area of<br />
the rectangles.<br />
28. One possible answer:<br />
Plot the speeds vs. time. Connect the points and find the area under the<br />
line graph. The definite integral also gives the area under the curve.<br />
39. (a, b) Take dm d dA as m k and letting dA → 0, k →∞in the center of<br />
mass equations.<br />
7. (a) Right: 0 t p/2, 3p/2 t 2p<br />
Left: p/2 t 3p/2<br />
Stopped: t p/2, 3p/2<br />
(b) 0; 3 (c) 2e (2/e) 4.7<br />
8. (a) Right: 0 t 3<br />
Left: never<br />
Stopped: t 0<br />
(b) (ln 10)/2 1.15; 4.15 (c) (ln 10)/2 1.15
390 Chapter 7 Applications of Definite Integrals<br />
7.2<br />
What you’ll learn about<br />
• Area Between Curves<br />
• Area Enclosed by Intersecting<br />
Curves<br />
• Boundaries with Changing<br />
Functions<br />
• Integrating with Respect to y<br />
• Saving Time with Geometric Formulas<br />
. . . and why<br />
The techniques of this section<br />
allow us to compute areas of<br />
complex regions of the plane.<br />
Areas in the Plane<br />
Area Between Curves<br />
We know how to find the area of a region between a curve and the x-axis but many times<br />
we want to know the area of a region that is bounded above by one curve, y f x, and<br />
below by another, y gx (Figure 7.3).<br />
We find the area as an integral by applying the first two steps of the modeling strategy<br />
developed in Section 7.1.<br />
1. We partition the region into vertical strips of equal width Δx and approximate each<br />
strip with a rectangle with base parallel to a, b (Figure 7.4). Each rectangle has area<br />
f c k gc k Δx<br />
for some c k in its respective subinterval (Figure 7.5). This expression will be nonnegative<br />
even if the region lies below the x-axis. We approximate the area of the region<br />
with the Riemann sum<br />
f c k gc k Δx.<br />
y<br />
Upper curve<br />
y f(x)<br />
y<br />
y f(x)<br />
y<br />
(c k , f (c k ))<br />
f (c k ) g(c k )<br />
a<br />
b<br />
x<br />
a<br />
b<br />
x<br />
a<br />
b<br />
x<br />
c k<br />
x<br />
Lower curve<br />
y g(x)<br />
y g(x)<br />
(c k , g(c k ))<br />
Figure 7.3 The region between y f (x)<br />
and y g(x) and the lines x a and<br />
x b.<br />
Figure 7.4 We approximate the region<br />
with rectangles perpendicular to the x-axis.<br />
2. The limit of these sums as Δx→0 is<br />
b<br />
f x gx dx.<br />
a<br />
Figure 7.5 The area of a typical rectangle<br />
is f (c k ) g(c k ) Δx.<br />
This approach to finding area captures the properties of area, so it can serve as a<br />
definition.<br />
DEFINITION<br />
Area Between Curves<br />
If f and g are continuous with f x gx throughout a, b, then the area between<br />
the curves y f (x) and y g(x) from a to b is the integral of f g from a to b,<br />
A b<br />
a<br />
f x gx dx.
Section 7.2 Areas in the Plane 391<br />
y<br />
2<br />
y sec 2 x<br />
EXAMPLE 1<br />
Applying the Definition<br />
Find the area of the region between y sec 2 x and y sin x from x 0 to x p4.<br />
SOLUTION<br />
We graph the curves (Figure 7.6) to find their relative positions in the plane, and see that<br />
y sec 2 x lies above y sin x on 0, p4. The area is therefore<br />
1<br />
y sin x<br />
A p4<br />
sec 2 x sin x dx<br />
0<br />
p4<br />
<br />
[ ]<br />
tan x cos x 0<br />
2<br />
units squared.<br />
2<br />
0<br />
–4<br />
x<br />
Now try Exercise 1.<br />
Figure 7.6 The region in Example 1.<br />
y 1 = 2k – k sin kx<br />
y 2 = k sin kx<br />
k = 2<br />
[0, ] by [0, 4]<br />
k = 1<br />
Figure 7.7 Two members of the family<br />
of butterfly-shaped regions described in<br />
Exploration 1.<br />
EXPLORATION 1<br />
A Family of Butterflies<br />
For each positive integer k, let A k denote the area of the butterfly-shaped region enclosed<br />
between the graphs of y k sin kx and y 2k k sin kx on the interval<br />
0, pk. The regions for k 1 and k 2 are shown in Figure 7.7.<br />
1. Find the areas of the two regions in Figure 7.7.<br />
2. Make a conjecture about the areas A k for k 3.<br />
3. Set up a definite integral that gives the area A k . Can you make a simple<br />
u-substitution that will transform this integral into the definite integral that<br />
gives the area A 1 ?<br />
4. What is lim k→∞ A k ?<br />
5. If P k denotes the perimeter of the kth butterfly-shaped region, what is<br />
lim k→∞ P k ? (You can answer this question without an explicit formula for P k .)<br />
y 1 = 2 – x 2<br />
y 2 = – x<br />
[–6, 6] by [–4, 4]<br />
Figure 7.8 The region in Example 2.<br />
Area Enclosed by Intersecting Curves<br />
When a region is enclosed by intersecting curves, the intersection points give the limits of<br />
integration.<br />
EXAMPLE 2<br />
Area of an Enclosed Region<br />
Find the area of the region enclosed by the parabola y 2 x 2 and the line y x.<br />
SOLUTION<br />
We graph the curves to view the region (Figure 7.8).<br />
The limits of integration are found by solving the equation<br />
2 x 2 x<br />
either algebraically or by calculator. The solutions are x 1 and x 2.<br />
continued
392 Chapter 7 Applications of Definite Integrals<br />
Since the parabola lies above the line on 1, 2, the area integrand is 2 x 2 x.<br />
A 2<br />
1<br />
2 x 2 x dx<br />
[ 2x x x 3 2<br />
3<br />
2<br />
]<br />
2<br />
1<br />
9 2 units squared Now try Exercise 5.<br />
y 1 = 2 cos x<br />
y 2 = x 2 – 1<br />
[–3, 3] by [–2, 3]<br />
Figure 7.9 The region in Example 3.<br />
Finding Intersections by<br />
Calculator<br />
The coordinates of the points of intersection<br />
of two curves are sometimes<br />
needed for other calculations. To take<br />
advantage of the accuracy provided by<br />
calculators, use them to solve for the<br />
values and store the ones you want.<br />
EXAMPLE 3<br />
Using a Calculator<br />
Find the area of the region enclosed by the graphs of y 2 cos x and y x 2 1.<br />
SOLUTION<br />
The region is shown in Figure 7.9.<br />
Using a calculator, we solve the equation<br />
2 cos x x 2 1<br />
to find the x-coordinates of the points where the curves intersect. These are the limits of<br />
integration. The solutions are x 1.265423706. We store the negative value as A and<br />
the positive value as B. The area is<br />
NINT 2 cos x x 2 1, x, A, B4.994907788.<br />
This is the final calculation, so we are now free to round. The area is about 4.99.<br />
Now try Exercise 7.<br />
Boundaries with Changing Functions<br />
If a boundary of a region is defined by more than one function, we can partition the region<br />
into subregions that correspond to the function changes and proceed as usual.<br />
EXAMPLE 4<br />
Finding Area Using Subregions<br />
Find the area of the region R in the first quadrant that is bounded above by y x and<br />
below by the x-axis and the line y x 2.<br />
SOLUTION<br />
The region is shown in Figure 7.10.<br />
2<br />
y<br />
4<br />
⌠<br />
Area ⎡⎣<br />
⎮ √⎯⎯x x 2 dx<br />
⌡<br />
2<br />
2<br />
⌠<br />
Area ⎮ √⎯⎯x dx<br />
⌡0<br />
⎣<br />
⎡<br />
y ⎯√⎯x<br />
(4, 2)<br />
1<br />
B<br />
y x 2<br />
A<br />
0<br />
y 0 2<br />
4<br />
x<br />
Figure 7.10 Region R split into subregions A and B. (Example 4)<br />
continued
Section 7.2 Areas in the Plane 393<br />
While it appears that no single integral can give the area of R (the bottom boundary is<br />
defined by two different curves), we can split the region at x 2 into two regions A and<br />
B. The area of R can be found as the sum of the areas of A and B.<br />
Area of R 2<br />
x dx 4<br />
x x 2 dx<br />
0<br />
2<br />
x32]<br />
area of A<br />
area of B<br />
2 2<br />
<br />
3 [ 2 3 x32 x 4<br />
2<br />
<br />
2<br />
2x]<br />
0 2<br />
1 0<br />
units squared<br />
3<br />
Now try Exercise 9.<br />
Integrating with Respect to y<br />
Sometimes the boundaries of a region are more easily described by functions of y than by<br />
functions of x. We can use approximating rectangles that are horizontal rather than vertical<br />
and the resulting basic formula has y in place of x.<br />
For regions like these<br />
y<br />
y<br />
y<br />
d<br />
x f(y)<br />
d<br />
x f(y)<br />
d<br />
x g(y)<br />
x f(y)<br />
x g(y)<br />
c<br />
x g(y)<br />
c<br />
0<br />
c<br />
x<br />
0<br />
x<br />
0<br />
x<br />
use this formula<br />
d<br />
A =<br />
∫<br />
[f (y) – g(y)]dy.<br />
c<br />
2<br />
1<br />
0<br />
y<br />
(g(y), y)<br />
y 0<br />
y<br />
f(y) g(y)<br />
x y 2 (4, 2)<br />
( f (y), y)<br />
x y 2<br />
2 4<br />
Figure 7.11 It takes two integrations to<br />
find the area of this region if we integrate<br />
with respect to x. It takes only one if we<br />
integrate with respect to y. (Example 5)<br />
x<br />
EXAMPLE 5<br />
Integrating with Respect to y<br />
Find the area of the region in Example 4 by integrating with respect to y.<br />
SOLUTION<br />
We remarked in solving Example 4 that “it appears that no single integral can give the<br />
area of R,” but notice how appearances change when we think of our rectangles being<br />
summed over y. The interval of integration is 0, 2, and the rectangles run between<br />
the same two curves on the entire interval. There is no need to split the region<br />
(Figure 7.11).<br />
We need to solve for x in terms of y in both equations:<br />
y x 2 becomes x y 2,<br />
y x becomes x y 2 , y 0.<br />
continued
394 Chapter 7 Applications of Definite Integrals<br />
y 1 = x 3 , y 2 = √x + 2, y 3 = – √x + 2<br />
(c, d)<br />
We must still be careful to subtract the lower number from the higher number when forming<br />
the integrand. In this case, the higher numbers are the higher x-values, which are on<br />
the line x y 2 because the line lies to the right of the parabola. So,<br />
Area of R 2<br />
0<br />
y 2 y 2 dy <br />
[ y 2y y 2 3<br />
2<br />
3<br />
]<br />
2<br />
1 0<br />
units squared.<br />
3<br />
0<br />
Now try Exercise 11.<br />
(a, b)<br />
[–3, 3] by [–3, 3]<br />
Figure 7.12 The region in Example 6.<br />
(a, b)<br />
[–3, 3] by [–3, 3]<br />
(c, d)<br />
Figure 7.13 If we integrate with<br />
respect to x in Example 6, we must split<br />
the region at x a.<br />
EXAMPLE 6<br />
Making the Choice<br />
Find the area of the region enclosed by the graphs of y x 3 and x y 2 2.<br />
SOLUTION<br />
We can produce a graph of the region on a calculator by graphing the three curves<br />
y x 3 , y x 2, and y x 2 (Figure 7.12).<br />
This conveniently gives us all of our bounding curves as functions of x. If we integrate<br />
in terms of x, however, we need to split the region at x a (Figure 7.13).<br />
On the other hand, we can integrate from y b to y d and handle the entire region<br />
at once. We solve the cubic for x in terms of y:<br />
y x 3 becomes x y 13 .<br />
To find the limits of integration, we solve y 13 y 2 2. It is easy to see that the<br />
lower limit is b 1, but a calculator is needed to find that the upper limit<br />
d 1.793003715. We store this value as D.<br />
The cubic lies to the right of the parabola, so<br />
Area NINT y 13 y 2 2, y, 1, D 4.214939673.<br />
The area is about 4.21. Now try Exercise 27.<br />
Saving Time with Geometry Formulas<br />
Here is yet another way to handle Example 4.<br />
2<br />
1<br />
0<br />
y<br />
y 0<br />
y √⎯x<br />
y x 2 2<br />
Area 2<br />
2<br />
2<br />
4<br />
(4, 2)<br />
Figure 7.14 The area of the blue region<br />
is the area under the parabola y x<br />
minus the area of the triangle. (Example 7)<br />
x<br />
EXAMPLE 7<br />
Using Geometry<br />
Find the area of the region in Example 4 by subtracting the area of the triangular region<br />
from the area under the square root curve.<br />
SOLUTION<br />
Figure 7.14 illustrates the strategy, which enables us to integrate with respect to x without<br />
splitting the region.<br />
Area x32]<br />
4<br />
x dx 1 4<br />
2 22 2 2 1 0<br />
units squared<br />
3 3<br />
0<br />
0<br />
Now try Exercise 35.<br />
The moral behind Examples 4, 5, and 7 is that you often have options for finding the<br />
area of a region, some of which may be easier than others. You can integrate with respect to<br />
x or with respect to y, you can partition the region into subregions, and sometimes you can<br />
even use traditional geometry formulas. Sketch the region first and take a moment to determine<br />
the best way to proceed.
Section 7.2 Areas in the Plane 395<br />
Quick Review 7.2 (For help, go to Sections 1.2 and 5.1.)<br />
In Exercises 1–5, find the area between the x-axis and the graph of<br />
the given function over the given interval.<br />
1. y sin x over 0, p 2<br />
2. y e 2x over 0, 1 1 2 (e 2 1) 3.195<br />
3. y sec 2 x over p4, p4 2<br />
4. y 4x x 3 over 0, 2 4<br />
5. y 9 x 2 over 3, 3 9p/2<br />
In Exercises 6–10, find the x- and y-coordinates of all points where<br />
the graphs of the given functions intersect. If the curves never<br />
intersect, write “NI.”<br />
6. y x 2 4x and y x 6 (6, 12); (1, 5)<br />
7. y e x and y x 1 (0, 1)<br />
8. y x 2 px and y sin x (0, 0); (p, 0)<br />
2x<br />
9. y x<br />
2<br />
and y x<br />
1<br />
3 (1, 1); (0, 0); (1, 1)<br />
10. y sin x and y x 3<br />
(–0.9286, –0.8008); (0, 0); (0.9286, 0.8008)<br />
Section 7.2 Exercises<br />
In Exercises 1–6, find the area of the shaded region analytically.<br />
1. y<br />
p/2<br />
4. y<br />
4/3<br />
1 x 12y 2 12y 3<br />
1 y 1<br />
y cos 2 x<br />
x<br />
0 –2<br />
<br />
x 2y 2 2y<br />
x<br />
0<br />
1<br />
5. y<br />
128/15<br />
2. y<br />
4p/3<br />
y<br />
1<br />
– sec 2 t 2<br />
2<br />
(–2, 8)<br />
8<br />
(2, 8)<br />
–3 –<br />
1<br />
0<br />
–3<br />
t<br />
y x 4 2x 2<br />
y 4 sin 2 t<br />
–2<br />
–1<br />
–1<br />
y 2x 2 2<br />
1<br />
x<br />
–4<br />
NOT TO SCALE<br />
6. y<br />
22/15<br />
3. y<br />
1/12<br />
1<br />
y x 2<br />
1<br />
x y 3<br />
(1, 1)<br />
–1<br />
0<br />
1<br />
x<br />
x y 2<br />
0 1<br />
x<br />
–2<br />
y –2x 4
396 Chapter 7 Applications of Definite Integrals<br />
In Exercises 7 and 8, use a calculator to find the area of the region<br />
enclosed by the graphs of the two functions.<br />
7. y sin x, y 1 x 2 1.670 8. y cos(2x), y x 2 2 4.332<br />
In Exercises 9 and 10, find the area of the shaded region analytically.<br />
9. y<br />
5/6<br />
y x<br />
1<br />
0<br />
10. y<br />
5/6<br />
1<br />
0 1<br />
In Exercises 11 and 12, find the area enclosed by the graphs of the<br />
two curves by integrating with respect to y.<br />
11. y 2 x 1, y 2 3 x 12. y 2 x 3, y 2x<br />
7.542 7.146<br />
In Exercises 13 and 14, find the total shaded area.<br />
13. y<br />
16<br />
y x 2 3x<br />
2<br />
(2, 2)<br />
–2<br />
14. y<br />
8 1 6 <br />
(–2, 4)<br />
–1<br />
(–2, –10)<br />
–10<br />
4<br />
2<br />
–2 –1 1 2<br />
–5<br />
1<br />
y 1<br />
x 2<br />
y — 4<br />
1<br />
2<br />
x y 2<br />
y x 2 2<br />
2<br />
x<br />
y 2x 3 x 2 5x<br />
y 4 x 2<br />
x<br />
3<br />
y x 2<br />
(3, –5)<br />
x<br />
x<br />
In Exercises 15–34, find the area of the regions enclosed by the lines<br />
and curves.<br />
15. y x 2 2 and y 2 10 2 3 <br />
16. y 2x x 2 and y 3 10 2 3 <br />
17. y 7 2x 2 and y x 2 4 4<br />
18. y x 4 4x 2 4 and y x 2 8<br />
19. y x a 2 x , 2 a 0, and y 0 2 3 a 3<br />
20. y x and 5y x 6 1 2 3 (3 points of intersection)<br />
(How many intersection points are there?)<br />
21. y x 2 4 and y x 2 2 4 21 1 3 <br />
22. x y 2 and x y 2 4 1 2 <br />
23. y 2 4x 4 and 4x y 16 30 3 8 <br />
24. x y 2 0 and x 2y 2 3 4<br />
25. x y 2 0 and x 3y 2 2 8/3<br />
26. 4x 2 y 4 and x 4 y 1 6 1 1<br />
27. x y 2 3 and 4x y 2 0 8<br />
28. y 2 sin x and y sin 2x, 0 x p 4<br />
29. y 8 cos x and y sec 2 x, p3 x p3 63<br />
30. y cos px2 and y 1 x 2 4 3 4 0.0601<br />
p<br />
31. y sin px2 and y x 4 p<br />
0.273<br />
p<br />
32. y sec 2 x, y tan 2 x, x p4, x p4 p 2 <br />
33. x tan 2 y and x tan 2 y, p4 y p4 4 p 0.858<br />
34. x 3 sin y cosy and x 0, 0 y p2 2<br />
In Exercises 35 and 36, find the area of the region by subtracting the<br />
area of a triangular region from the area of a larger region.<br />
35. The region on or above the x-axis bounded by the curves<br />
y 2 x 3 and y 2x. 4.333<br />
36. The region on or above the x-axis bounded by the curves<br />
y 4 x 2 and y 3x. 152<br />
37. Find the area of the propeller-shaped region enclosed by<br />
the curve x y 3 0 and the line x y 0. 1/2<br />
38. Find the area of the region in the first quadrant bounded<br />
by the line y x, the line x 2, the curve y 1x 2 , and the<br />
x-axis. 1<br />
39. Find the area of the “triangular” region in the first quadrant<br />
bounded on the left by the y-axis and on the right by the curves<br />
y sin x and y cos x. 2 1 0.414<br />
40. Find the area of the region between the curve y 3 x 2 and<br />
the line y 1 by integrating with respect to (a) x, (b) y. 32/3<br />
41. The region bounded below by the parabola y x 2 and above by<br />
the line y 4 is to be partitioned into two subsections of equal<br />
area by cutting across it with the horizontal line y c.<br />
(a) Sketch the region and draw a line y c across it that<br />
looks about right. In terms of c, what are the coordinates of<br />
the points where the line and parabola intersect? Add them to<br />
your figure. (–c, c); (c, c)<br />
(b) Find c by integrating with respect to y. (This puts c in the<br />
limits of integration.) y dy 4<br />
y dy ⇒ c 2 43<br />
c<br />
0<br />
(c) Find c by integrating with respect to x. (This puts c into the<br />
integrand as well.)<br />
42. Find the area of the region in the first quadrant bounded on the<br />
left by the y-axis, below by the line y x4, above left by the<br />
curve y 1 x, and above right by the curve y 2x. 11/3<br />
41. (c) c<br />
(c x 2 ) dx (4 c)c 2 (4 x 2 ) dx ⇒ c 2 43<br />
0<br />
c<br />
c<br />
4<br />
<br />
5
Section 7.2 Areas in the Plane 397<br />
43. The figure here shows triangle AOC inscribed in the region cut<br />
from the parabola y x 2 by the line y a 2 . Find the limit of<br />
the ratio of the area of the triangle to the area of the parabolic<br />
region as a approaches zero. 3/4<br />
44. Suppose the area of the region between the graph of a positive<br />
continuous function f and the x-axis from x a to x b is<br />
4 square units. Find the area between the curves y f x and<br />
y 2 f x from x a to x b. 4<br />
45. Writing to Learn Which of the following integrals, if either,<br />
calculates the area of the shaded region shown here? Give<br />
reasons for your answer. Neither; both are zero<br />
i. 1<br />
x x dx 1<br />
2xdx<br />
1<br />
1<br />
ii. 1<br />
x x dx 1<br />
2xdx<br />
1<br />
A<br />
(–a, a 2 )<br />
–a<br />
–1<br />
1<br />
y – x<br />
O<br />
1<br />
–1<br />
y<br />
y<br />
y x 2<br />
y x<br />
C y a 2<br />
(a, a 2 )<br />
a<br />
1<br />
x<br />
x<br />
Standardized Test Questions<br />
You should solve the following problems without using a<br />
graphing calculator.<br />
50. True or False The area of the region enclosed by the graph of<br />
y x 2 1 and the line y 10 is 36. Justify your answer.<br />
True. 36 is the value of the appropriate integral.<br />
51. True or False The area of the region in the first quadrant<br />
enclosed by the graphs of y cos x, y x, and the y-axis is<br />
given by the definite integral 0.739<br />
(x cos x) dx. Justify your<br />
0<br />
answer.<br />
False. It is 0.739<br />
(cos x x) dx.<br />
0<br />
52. Multiple Choice Let R be the region in the first quadrant<br />
bounded by the x-axis, the graph of x y 2 2, and the line<br />
x 4. Which of the following integrals gives the area of R?<br />
2<br />
(A)0<br />
(C)2<br />
2<br />
[4 (y 2 2)]dy (B) [(y 2 2) 4]dy<br />
0<br />
2<br />
[4 (y 2 2)]dy (D) [(y 2 2) 4]dy<br />
2<br />
4<br />
(E) [4 (y 2 2)]dy<br />
2<br />
53. Multiple Choice Which of the following gives the area of the<br />
region between the graphs of y x 2 and y x from x 0 to<br />
x 3? E<br />
(A) 2 (B) 9/2 (C) 13/2 (D) 13 (E) 27/2<br />
54. Multiple Choice Let R be the shaded region enclosed by the<br />
graphs of y e x2 , y sin(3x), and the y-axis as shown in the<br />
figure below. Which of the following gives the approximate area<br />
of the region R? B<br />
(A) 1.139 (B) 1.445 (C) 1.869 (D) 2.114 (E) 2.340<br />
2<br />
y<br />
A<br />
46. Writing to Learn Is the following statement true, sometimes<br />
true, or never true? The area of the region between the graphs<br />
of the continuous functions y f x and y gx and the<br />
vertical lines x a and x b (a b) is<br />
b<br />
f x gx dx.<br />
a<br />
Sometimes; If f (x) g(x) on (a, b),<br />
Give reasons for your answer.<br />
then true.<br />
47. Find the area of the propeller-shaped region enclosed between<br />
the graphs of ln 4 (1/2) 0.886<br />
2x<br />
y x<br />
2<br />
and y x<br />
1<br />
3 .<br />
48. Find the area of the propeller-shaped region enclosed between<br />
the graphs of y sin x and y x 3 . 0.4303<br />
49. Find the positive value of k such that the area of the region<br />
enclosed between the graph of y k cos x and the graph of<br />
y kx 2 is 2. k 1.8269<br />
55. Multiple Choice Let f and g be the functions given by<br />
f (x) e x and g(x) 1/x. Which of the following gives the area<br />
of the region enclosed by the graphs of f and g between x 1<br />
and x 2? A<br />
(A) e 2 e ln2<br />
(B) ln 2 e 2 e<br />
(C) e 2 1 2 <br />
(D) e 2 e 1 2 <br />
(E) 1 e ln2<br />
0<br />
– 2<br />
2<br />
x
398 Chapter 7 Applications of Definite Integrals<br />
Exploration<br />
56. Group Activity Area of Ellipse<br />
An ellipse with major axis of length 2a and minor axis of<br />
length 2b can be coordinatized with its center at the origin and<br />
its major axis horizontal, in which case it is defined by the<br />
equation<br />
Extending the Ideas<br />
57. Cavalieri’s Theorem Bonaventura Cavalieri (1598–1647)<br />
discovered that if two plane regions can be arranged to lie over<br />
the same interval of the x-axis in such a way that they have<br />
identical vertical cross sections at every point (see figure), then<br />
the regions have the same area. Show that this theorem is true.<br />
x<br />
2 a<br />
2 y 2 b<br />
2 1.<br />
(a) Find the equations that define the upper and lower<br />
semiellipses as functions of x.<br />
(b) Write an integral expression that gives the area of the ellipse.<br />
(c) With your group, use NINT to find the areas of ellipses for<br />
various lengths of a and b.<br />
(d) There is a simple formula for the area of an ellipse with<br />
major axis of length 2a and minor axis of length 2b. Can you<br />
tell what it is from the areas you and your group have found?<br />
(e) Work with your group to write a proof of this area formula<br />
by showing that it is the exact value of the integral expression<br />
in part (b).<br />
<br />
56. (a) y b <br />
1 a<br />
x 2 2 <br />
(b) 2 a b<br />
a <br />
x<br />
1 <br />
2 a<br />
2 dx<br />
(c) Answers may vary.<br />
(d, e) abp<br />
a x b<br />
Cross sections have<br />
the same length at<br />
every point in [a, b].<br />
58. Find the area of the region enclosed by the curves<br />
x<br />
y x 2 and<br />
1<br />
y mx, 0 m 1.<br />
m ln (m) 1<br />
57. Since f (x) g(x) is the same for each region where f (x) and g(x) represent<br />
the upper and lower edges, area b [ f (x) g(x)] dx will be the same for<br />
a<br />
each.
Section 7.3 Volumes 399<br />
7.3<br />
Volumes<br />
What you’ll learn about<br />
• Volume As an Integral<br />
• Square Cross Sections<br />
• Circular Cross Sections<br />
• Cylindrical Shells<br />
• Other Cross Sections<br />
. . . and why<br />
The techniques of this section<br />
allow us to compute volumes of<br />
certain solids in three dimensions.<br />
y<br />
S<br />
P x<br />
Cross-section R(x)<br />
with area A(x)<br />
Volume As an Integral<br />
In Section 5.1, Example 3, we estimated the volume of a sphere by partitioning it into thin<br />
slices that were nearly cylindrical and summing the cylinders’ volumes using MRAM.<br />
MRAM sums are Riemann sums, and had we known how at the time, we could have continued<br />
on to express the volume of the sphere as a definite integral.<br />
Starting the same way, we can now find the volumes of a great many solids by integration.<br />
Suppose we want to find the volume of a solid like the one in Figure 7.15. The cross<br />
section of the solid at each point x in the interval a, b is a region Rx of area Ax. If A is a<br />
continuous function of x, we can use it to define and calculate the volume of the solid as an<br />
integral in the following way.<br />
We partition a, b into subintervals of length Δx and slice the solid, as we would a loaf<br />
of bread, by planes perpendicular to the x-axis at the partition points. The kth slice, the<br />
one between the planes at x k1 and x k , has approximately the same volume as the cylinder<br />
between the two planes based on the region Rx k (Figure 7.16).<br />
y<br />
Plane at x k1<br />
Approximating<br />
cylinder based<br />
on R(x k ) has height<br />
Δx k x k x k1<br />
0<br />
a<br />
x<br />
0<br />
x k1<br />
Plane at x k<br />
Figure 7.15 The cross section of an<br />
arbitrary solid at point x.<br />
b<br />
x<br />
The cylinder’s base<br />
is the region R(x k )<br />
with area A(x k )<br />
NOT TO SCALE<br />
x k<br />
x<br />
Figure 7.16 Enlarged view of the slice of the solid between the planes at x k1 and x k .<br />
The volume of the cylinder is<br />
V k base area height Ax k Δx.<br />
The sum<br />
V k Ax k Δx<br />
approximates the volume of the solid.<br />
This is a Riemann sum for Ax on a, b. We expect the approximations to improve as<br />
the norms of the partitions go to zero, so we define their limiting integral to be the volume<br />
of the solid.<br />
DEFINITION<br />
Volume of a Solid<br />
The volume of a solid of known integrable cross section area Ax from x a to<br />
x b is the integral of A from a to b,<br />
V b<br />
a<br />
Ax dx.
400 Chapter 7 Applications of Definite Integrals<br />
To apply the formula in the previous definition, we proceed as follows.<br />
y<br />
How to Find Volume by the Method of Slicing<br />
1. Sketch the solid and a typical cross section.<br />
2. Find a formula for Ax.<br />
3. Find the limits of integration.<br />
4. Integrate Ax to find the volume.<br />
Typical cross-section<br />
0<br />
x<br />
x<br />
x<br />
x (m)<br />
Figure 7.17 A cross section of the pyramid<br />
in Example 1.<br />
3<br />
3<br />
3<br />
Square Cross Sections<br />
Let us apply the volume formula to a solid with square cross sections.<br />
EXAMPLE 1<br />
A Square-Based Pyramid<br />
A pyramid 3 m high has congruent triangular sides and a square base that is 3 m on each<br />
side. Each cross section of the pyramid parallel to the base is a square. Find the volume<br />
of the pyramid.<br />
SOLUTION<br />
We follow the steps for the method of slicing.<br />
1. Sketch. We draw the pyramid with its vertex at the origin and its altitude along the<br />
interval 0 x 3. We sketch a typical cross section at a point x between 0 and 3<br />
(Figure 7.17).<br />
2. Find a formula for Ax. The cross section at x is a square x meters on a side, so<br />
Ax x 2 .<br />
3. Find the limits of integration. The squares go from x 0 to x 3.<br />
4. Integrate to find the volume.<br />
V 3<br />
Ax dx 3<br />
x 2 x 3<br />
3<br />
3<br />
] 9m 3<br />
0<br />
0<br />
0<br />
Now try Exercise 3.<br />
[–3, 3] by [–4, 4]<br />
Figure 7.18 The region in Example 2.<br />
y<br />
f(x)<br />
Figure 7.19 The region in Figure 7.18<br />
is revolved about the x-axis to generate a<br />
solid. A typical cross section is circular,<br />
with radius f (x) 2 x cos x.<br />
(Example 2)<br />
x<br />
Circular Cross Sections<br />
The only thing that changes when the cross sections of a solid are circular is the formula<br />
for Ax. Many such solids are solids of revolution, as in the next example.<br />
EXAMPLE 2<br />
A Solid of Revolution<br />
The region between the graph of f x 2 x cos x and the x-axis over the interval<br />
2, 2 is revolved about the x-axis to generate a solid. Find the volume of the solid.<br />
SOLUTION<br />
Revolving the region (Figure 7.18) about the x-axis generates the vase-shaped solid in<br />
Figure 7.19. The cross section at a typical point x is circular, with radius equal to f x.<br />
Its area is<br />
Ax p f x 2 .<br />
continued
Section 7.3 Volumes 401<br />
The volume of the solid is<br />
V 2<br />
2<br />
Ax dx<br />
NINT p2 x cos x 2 , x, 2, 2 52.43 units cubed.<br />
Now try Exercise 7.<br />
[–/4, /2] by [–1.5, 1.5]<br />
EXAMPLE 3<br />
Washer Cross Sections<br />
The region in the first quadrant enclosed by the y-axis and the graphs of y cos x and<br />
y sin x is revolved about the x-axis to form a solid. Find its volume.<br />
SOLUTION<br />
The region is shown in Figure 7.20.<br />
We revolve it about the x-axis to generate a solid with a cone-shaped cavity in its center<br />
(Figure 7.21).<br />
Figure 7.20 The region in Example 3.<br />
Figure 7.22 The area of a washer is<br />
pR 2 pr 2 . (Example 3)<br />
CAUTION!<br />
The area of a washer is pR 2 pr 2 ,<br />
which you can simplify to p(R 2 r 2 ),<br />
but not to p(R r) 2 . No matter how<br />
tempting it is to make the latter simplification,<br />
it’s wrong. Don’t do it.<br />
r<br />
R<br />
Figure 7.21 The solid generated by revolving the region<br />
in Figure 7.20 about the x-axis. A typical cross section is<br />
a washer: a circular region with a circular region cut out of<br />
its center. (Example 3)<br />
This time each cross section perpendicular to the axis of revolution is a washer, a circular<br />
region with a circular region cut from its center. The area of a washer can be found<br />
by subtracting the inner area from the outer area (Figure 7.22).<br />
In our region the cosine curve defines the outer radius, and the curves intersect at<br />
x p4. The volume is<br />
V p4<br />
0<br />
p p4<br />
0<br />
pcos 2 x sin 2 x dx<br />
cos 2x dx<br />
]<br />
p[ sin 2x<br />
2<br />
p4<br />
0<br />
identity: cos 2 x sin 2 x cos 2x<br />
p 2 units cubed. Now try Exercise 17.<br />
We could have done the integration in Example 3 with NINT, but we wanted to demonstrate<br />
how a trigonometric identity can be useful under unexpected circumstances in calculus.<br />
The double-angle identity turned a difficult integrand into an easy one and enabled us to<br />
get an exact answer by antidifferentiation.<br />
Cylindrical Shells<br />
There is another way to find volumes of solids of rotation that can be useful when the axis<br />
of revolution is perpendicular to the axis containing the natural interval of integration. Instead<br />
of summing volumes of thin slices, we sum volumes of thin cylindrical shells that<br />
grow outward from the axis of revolution like tree rings.
402 Chapter 7 Applications of Definite Integrals<br />
[–6, 4] by [–3, 3]<br />
Figure 7.23 The graph of the region in<br />
Exploration 1, before revolution.<br />
Axis of<br />
revolution<br />
x –1<br />
Figure 7.24 The region in Figure 7.23 is<br />
revolved about the line x 1 to form a<br />
solid cake. The natural interval of integration<br />
is along the x-axis, perpendicular to<br />
the axis of revolution. (Exploration 1)<br />
0<br />
y<br />
y<br />
0 x k<br />
y k<br />
Figure 7.25 Cutting the cake into<br />
thin cylindrical slices, working from the<br />
inside out. Each slice occurs at some x k<br />
between 0 and 3 and has thickness Δx.<br />
(Exploration 1)<br />
3<br />
3<br />
x<br />
EXPLORATION 1<br />
x<br />
Volume by Cylindrical Shells<br />
The region enclosed by the x-axis and the parabola y f x 3x x 2 is revolved<br />
about the line x 1 to generate the shape of a cake (Figures 7.23, 7.24). (Such a<br />
cake is often called a bundt cake.) What is the volume of the cake?<br />
Integrating with respect to y would be awkward here, as it is not easy to get the<br />
original parabola in terms of y. (Try finding the volume by washers and you will<br />
soon see what we mean.) To integrate with respect to x, you can do the problem by<br />
cylindrical shells, which requires that you cut the cake in a rather unusual way.<br />
1. Instead of cutting the usual wedge shape, cut a cylindrical slice by cutting<br />
straight down all the way around close to the inside hole. Then cut another<br />
cylindrical slice around the enlarged hole, then another, and so on. The radii of<br />
the cylinders gradually increase, and the heights of the cylinders follow the<br />
contour of the parabola: smaller to larger, then back to smaller (Figure 7.25).<br />
Each slice is sitting over a subinterval of the x-axis of length Δx. Its radius is<br />
approximately 1 x k . What is its height?<br />
2. If you unroll the cylinder at x k and flatten it out, it becomes (essentially) a rectangular<br />
slab with thickness Δx. Show that the volume of the slab is approximately<br />
2px k 13x k x 2 k Δx.<br />
3. 2px k 13x k x 2 k Δx is a Riemann sum. What is the limit of these Riemann<br />
sums as Δx→0?<br />
4. Evaluate the integral you found in step 3 to find the volume of the cake!<br />
EXAMPLE 4<br />
Finding Volumes Using Cylindrical Shells<br />
The region bounded by the curve y x, the x-axis, and the line x 4 is revolved<br />
about the x-axis to generate a solid. Find the volume of the solid.<br />
SOLUTION<br />
1. Sketch the region and draw a line segment across it parallel to the axis of revolution<br />
(Figure 7.26). Label the segment’s length (shell height) and distance from the axis of<br />
revolution (shell radius). The width of the segment is the shell thickness dy. (We<br />
drew the shell in Figure 7.27, but you need not do that.)<br />
Interval of<br />
integration<br />
y<br />
4 y 2<br />
Shell height<br />
2<br />
x y 2 (4, 2)<br />
⎧<br />
⎪<br />
y<br />
Shell<br />
⎨<br />
thickness dy<br />
⎪ y Shell radius<br />
⎩<br />
x<br />
0 4<br />
Figure 7.26 The region, shell dimensions, and<br />
interval of integration in Example 4.<br />
2<br />
y<br />
0<br />
y<br />
Shell height<br />
4 y 2<br />
y x<br />
(4, 2)<br />
Shell<br />
radius<br />
Figure 7.27 The shell swept out by the<br />
line segment in Figure 7.26.<br />
y<br />
x
Section 7.3 Volumes 403<br />
2. Identify the limits of integration: y runs from 0 to 2.<br />
3. Integrate to find the volume.<br />
V 2<br />
0<br />
2<br />
0<br />
2p( radius)( shell<br />
height) shell dy<br />
2py4 y 2 dy 8p<br />
Now try Exercise 33(a).<br />
4<br />
2<br />
y<br />
0 2<br />
Figure 7.28 The region and the height<br />
of a typical shell in Example 5.<br />
o<br />
x k<br />
[–1, 3.5] by [–0.8, 2.2]<br />
Figure 7.29 The base of the paperweight<br />
in Example 6. The segment perpendicular<br />
to the x-axis at x k is the diameter of<br />
a semicircle that is perpendicular to the<br />
base.<br />
0<br />
2<br />
y<br />
x<br />
x<br />
π<br />
y = 2 sin x<br />
Figure 7.30 Cross sections perpendicular<br />
to the region in Figure 7.29 are semicircular.<br />
(Example 6)<br />
<br />
EXAMPLE 5<br />
Finding Volumes Using Cylindrical Shells<br />
The region bounded by the curves y 4 x 2 , y x, and x 0 is revolved about the<br />
y-axis to form a solid. Use cylindrical shells to find the volume of the solid.<br />
SOLUTION<br />
1. Sketch the region and draw a line segment across it parallel to the y-axis<br />
(Figure 7.28). The segment’s length (shell height) is 4 x 2 x. The distance of the<br />
segment from the axis of revolution (shell radius) is x.<br />
2. Identify the limits of integration: The x-coordinate of the point of intersection of the<br />
curves y 4 x 2 and y x in the first quadrant is about 1.562. So x runs from 0 to<br />
1.562.<br />
3. Integrate to find the volume.<br />
Other Cross Sections<br />
radius)( height) shell dx<br />
1.562<br />
2p(x)(4 x 2 x) dx<br />
0<br />
13.327<br />
1.562<br />
V 2p ( shell<br />
0<br />
Now try Exercise 35.<br />
The method of cross-section slicing can be used to find volumes of a wide variety of unusually<br />
shaped solids, so long as the cross sections have areas that we can describe with<br />
some formula. Admittedly, it does take a special artistic talent to draw some of these<br />
solids, but a crude picture is usually enough to suggest how to set up the integral.<br />
EXAMPLE 6<br />
A Mathematician’s Paperweight<br />
A mathematician has a paperweight made so that its base is the shape of the region between<br />
the x-axis and one arch of the curve y 2 sin x (linear units in inches). Each cross<br />
section cut perpendicular to the x-axis (and hence to the xy-plane) is a semicircle whose<br />
diameter runs from the x-axis to the curve. (Think of the cross section as a semicircular<br />
fin sticking up out of the plane.) Find the volume of the paperweight.<br />
SOLUTION<br />
The paperweight is not easily drawn, but we know what it looks like. Its base is the region<br />
in Figure 7.29, and the cross sections perpendicular to the base are semicircular fins like<br />
those in Figure 7.30.<br />
The semicircle at each point x has<br />
radius 2s in x<br />
sin x and area Ax 1 2<br />
2 psin x2 .<br />
continued
404 Chapter 7 Applications of Definite Integrals<br />
The volume of the paperweight is<br />
V p<br />
0<br />
Ax dx<br />
p p<br />
2 sin x 2 dx<br />
0<br />
p 2 NINT sin x2 , x, 0,p<br />
p 2 1.570796327.<br />
Bonaventura Cavalieri<br />
(1598—1647)<br />
Cavalieri, a student of<br />
Galileo, discovered that<br />
if two plane regions<br />
can be arranged to lie<br />
over the same interval<br />
of the x-axis in such a<br />
way that they have<br />
identical vertical cross<br />
sections at every point, then the regions<br />
have the same area. This theorem and a<br />
letter of recommendation from Galileo<br />
were enough to win Cavalieri a chair at<br />
the University of Bologna in 1629. The<br />
solid geometry version in Example 7,<br />
which Cavalieri never proved, was<br />
named after him by later geometers.<br />
The number in parentheses looks like half of p, an observation that can be confirmed<br />
analytically, and which we support numerically by dividing by p to get 0.5. The volume<br />
of the paperweight is<br />
EXAMPLE 7<br />
2<br />
p 2 • p 2 p 2.47 in<br />
4<br />
3 .<br />
Cavalieri’s Volume Theorem<br />
Now try Exercise 39(a).<br />
Cavalieri’s volume theorem says that solids with equal altitudes and identical cross section<br />
areas at each height have the same volume (Figure 7.31). This follows immediately from<br />
the definition of volume, because the cross section area function Ax and the interval<br />
a, b are the same for both solids.<br />
b<br />
Same volume<br />
a<br />
Cross sections have<br />
the same length at<br />
every point in [a, b].<br />
Same cross-section<br />
area at every level<br />
Figure 7.31 Cavalieri’s volume theorem: These solids have the same volume. You can illustrate<br />
this yourself with stacks of coins. (Example 7)<br />
a x b<br />
Now try Exercise 43.
Section 7.3 Volumes 405<br />
EXPLORATION 2 Surface Area<br />
We know how to find the volume of a solid of revolution, but how would we find<br />
the surface area? As before, we partition the solid into thin slices, but now we wish<br />
to form a Riemann sum of approximations to surface areas of slices (rather than of<br />
volumes of slices).<br />
y<br />
y = f(x)<br />
x<br />
a<br />
b<br />
A typical slice has a surface area that can be approximated by 2p • f x • Δs,<br />
where Δs is the tiny slant height of the slice. We will see in Section 7.4, when<br />
we study arc length, that Δs Δx 2 , Δy 2 and that this can be written as<br />
Δs 1 f x k Δx.<br />
2<br />
Thus, the surface area is approximated by the Riemann sum<br />
n<br />
2p f x k 1 fx k 2 Δx.<br />
k1<br />
1. Write the limit of the Riemann sums as a definite integral from a to b. When<br />
will the limit exist?<br />
2. Use the formula from part 1 to find the surface area of the solid generated by<br />
revolving a single arch of the curve y sin x about the x-axis.<br />
3. The region enclosed by the graphs of y 2 x and x 4 is revolved about the<br />
x-axis to form a solid. Find the surface area of the solid.<br />
Quick Review 7.3 (For help, go to Section 1.2.)<br />
In Exercises 1–10, give a formula for the area of the plane region in 6. an isosceles right triangle with legs of length x x 2 /2<br />
terms of the single variable x.<br />
7. an isosceles right triangle with hypotenuse x x 2 /4<br />
1. a square with sides of length x x 2<br />
2. a square with diagonals of length x x 2 /2<br />
8. an isosceles triangle with two sides of length 2x<br />
and one side of length x (15/4)x 2<br />
3. a semicircle of radius x px 2 /2<br />
4. a semicircle of diameter x px 2 /8<br />
9. a triangle with sides 3x, 4x, and 5x 6x 2<br />
5. an equilateral triangle with sides of length x (3/4)x 2 10. a regular hexagon with sides of length x (33/2)x 2
406 Chapter 7 Applications of Definite Integrals<br />
Section 7.3 Exercises<br />
In Exercises 1 and 2, find a formula for the area A(x) of the cross<br />
sections of the solid that are perpendicular to the x-axis.<br />
1. The solid lies between planes perpendicular to the x-axis at<br />
x 1 and x 1. The cross sections perpendicular to<br />
the x-axis between these planes run from the semicircle<br />
y 1 x 2 to the semicircle y 1 x .<br />
2<br />
(a) The cross sections are circular disks with diameters in the<br />
xy-plane. p(1 x 2 )<br />
2. The solid lies between planes perpendicular to the x-axis at<br />
x 0 and x 4. The cross sections perpendicular to the x-axis<br />
between these planes run from y x to y x.<br />
(a) The cross sections are circular disks with diameters in the<br />
xy-plane. px<br />
y<br />
x 2 y 2 1<br />
y<br />
x y 2<br />
0<br />
4<br />
x<br />
–1<br />
1<br />
x<br />
(b) The cross sections are squares with bases in the xy-plane. 4x<br />
(b) The cross sections are squares with bases in the xy-plane. 4(1 x 2 )<br />
y<br />
y<br />
(c) The cross sections are squares with diagonals in the<br />
xy-plane. (The length of a square’s diagonal is 2 times<br />
the length of its sides.) 2(1 – x 2 )<br />
–1<br />
–1<br />
0<br />
x 2 y 2 1<br />
0<br />
1<br />
x 2 y 2 1<br />
1<br />
x<br />
y<br />
x<br />
x y 2<br />
(c) The cross sections are squares with diagonals in the<br />
xy-plane. 2x<br />
(d) The cross sections are equilateral triangles with bases<br />
in the xy-plane. 3x<br />
In Exercises 3–6, find the volume of the solid analytically.<br />
3. The solid lies between planes perpendicular to the x-axis<br />
at x 0 and x 4. The cross sections perpendicular to the<br />
axis on the interval 0 x 4 are squares whose diagonals run<br />
from y x to y x. 16<br />
4. The solid lies between planes perpendicular to the x-axis<br />
at x 1 and x 1. The cross sections perpendicular to the<br />
x-axis are circular disks whose diameters run from the parabola<br />
y x 2 to the parabola y 2 x 2 . 16p/15<br />
4<br />
x<br />
(d) The cross sections are equilateral triangles with bases in the<br />
xy-plane. 3(1 x 2 )<br />
2<br />
y<br />
y<br />
0<br />
y x 2<br />
–1<br />
0<br />
x 2 y 2 1<br />
1<br />
x<br />
y 2 x 2<br />
5. The solid lies between planes perpendicular to the x-axis at<br />
x 1 and x 1. The cross sections perpendicular to the<br />
x-axis between these planes are squares whose bases run from<br />
the semicircle y 1 x 2 to the semicircle y 1 x 2 .<br />
16/3<br />
x
Section 7.3 Volumes 407<br />
6. The solid lies between planes perpendicular to the x-axis<br />
at x 1 and x 1. The cross sections perpendicular<br />
to the x-axis between these planes are squares whose diagonals<br />
run from the semicircle y 1 x <br />
2 to the semicircle<br />
y 1 x . 2 8/3<br />
In Exercises 7–10, find the volume of the solid generated by<br />
revolving the shaded region about the given axis.<br />
7. about the x-axis 2p/3 8. about the y-axis 6p<br />
9. about the y-axis 4 p 10. about the x-axis p 2 /16<br />
0<br />
1<br />
1<br />
y<br />
y<br />
x 2y 2<br />
0 2<br />
x tan ⎛– y<br />
⎝ 4<br />
In Exercises 11–20, find the volume of the solid generated by<br />
revolving the region bounded by the lines and curves about the<br />
x-axis.<br />
11. y x 2 , y 0,<br />
32p/5<br />
x 2 12. y x 3 , y 0, x 2 128p/7<br />
13. y 9 x , 2 y 0 36p 14. y x x 2 , y 0 p/30<br />
15. y x, y 1, x 0 2p/3 16. y 2x, y x, x 1 p<br />
20. y x, y 2, x 0 8p<br />
In Exercises 21 and 22, find the volume of the solid generated by<br />
revolving the region about the given line.<br />
21. the region in the first quadrant bounded above by the line<br />
y 2, below by the curve y sec x tan x, and on the left by<br />
the y-axis, about the line y 2 2.301<br />
22. the region in the first quadrant bounded above by the line y 2,<br />
below by the curve y 2 sin x, 0 x p2, and on the left by<br />
the y-axis, about the line y 2 p(3p 8)<br />
In Exercises 23–28, find the volume of the solid generated by<br />
revolving the region about the y-axis.<br />
23. the region enclosed by x 5y 2 , x 0, y 1, y 1 2p<br />
24. the region enclosed by x y 32 , x 0, y 2 4p<br />
⎝<br />
⎛<br />
x<br />
25. the region enclosed by the triangle with vertices 1, 0, 2, 1,<br />
and 1, 1 4p/3<br />
26. the region enclosed by the triangle with vertices 0, 1, 1, 0,<br />
and 1, 1 2p/3<br />
x<br />
2<br />
y<br />
0 3<br />
0<br />
y<br />
x 3y/2<br />
y sin x cos x<br />
17. y x 2 1, y x 3 18. y 4 x 2 , y 2 x 108p/5<br />
117p/5<br />
19. y sec x, y 2, p4 x p4 p 2 2p<br />
x<br />
x<br />
27. the region in the first quadrant bounded above by the parabola<br />
y x 2 , below by the x-axis, and on the right by the line x 2<br />
8p<br />
28. the region bounded above by the curve y x and below by<br />
the line y x 2p/15<br />
Group Activity In Exercises 29–32, find the volume of the solid<br />
described.<br />
29. Find the volume of the solid generated by revolving the region<br />
bounded by y x and the lines y 2 and x 0 about<br />
(a) the x-axis. 8p (b) the y-axis. 32p/5<br />
(c) the line y 2. 8p/3 (d) the line x 4. 224p/15<br />
30. Find the volume of the solid generated by revolving the<br />
triangular region bounded by the lines y 2x, y 0, and<br />
x 1 about<br />
(a) the line x 1. 2p/3 (b) the line x 2. 8p/3<br />
31. Find the volume of the solid generated by revolving the region<br />
bounded by the parabola y x 2 and the line y 1 about<br />
(a) the line y 1. 16p/15 (b) the line y 2. 56p/15<br />
(c) the line y 1. 64p/15<br />
32. By integration, find the volume of the solid generated by<br />
revolving the triangular region with vertices 0, 0, b, 0,<br />
0, h about<br />
(a) the x-axis. (p/3)bh 2 (b) the y-axis. (p/3)b 2 h<br />
In Exercises 33 and 34, use the cylindrical shell method to find the<br />
volume of the solid generated by revolving the shaded region about<br />
the indicated axis.<br />
33. (a) the x-axis 6p/5 (b) the line y 1 4p/5<br />
(c) the line y 85 2p (d) the line y 25 2p<br />
y<br />
1 x 12(y 2 y 3 )<br />
0<br />
34. (a) the x-axis 8p/3 (b) the line y 2 8p/5<br />
(c) the line y 5 8p (d) the line y 58 4p<br />
y<br />
2<br />
0<br />
x y4 — 4<br />
y 2<br />
— 2<br />
y<br />
x 2<br />
— 2<br />
1<br />
1<br />
(2, 2)<br />
2<br />
x<br />
x
408 Chapter 7 Applications of Definite Integrals<br />
In Exercises 35–38, use the cylindrical shell method to find the<br />
volume of the solid generated by revolving the region bounded by<br />
the curves about the y-axis.<br />
35. y x, y x2, x 2 8p<br />
36. y x 2 , y 2 x, x 0, for x 0 5p/6<br />
37. y x, y 0, x 4 128p/5<br />
38. y 2x 1, y x, x 0 7p/15<br />
In Exercises 39–42, find the volume of the solid analytically.<br />
39. The base of a solid is the region between the curve<br />
y 2 si n x and the interval 0, p on the x-axis. The<br />
cross sections perpendicular to the x-axis are<br />
(a) equilateral triangles with bases running from the x-axis to<br />
the curve as shown in the figure. 23<br />
0<br />
2<br />
y 2<br />
(b) squares with bases running from the x-axis to the curve. 8<br />
40. The solid lies between planes perpendicular to the x-axis at<br />
x p3 and x p3. The cross sections perpendicular<br />
to the x-axis are<br />
(a) circular disks with diameters running from the curve<br />
y tan x to the curve y sec x. p3 (p 2 /6)<br />
(b) squares whose bases run from the curve y tan x to the<br />
curve y sec x. 43 (2p/3)<br />
41. The solid lies between planes perpendicular to the y-axis<br />
at y 0 and y 2. The cross sections perpendicular to the<br />
y-axis are circular disks with diameters running from the<br />
y-axis to the parabola x 5y 2 . 8p<br />
42. The base of the solid is the disk x 2 y 2 1. The cross<br />
sections by planes perpendicular to the y-axis between<br />
y 1 and y 1 are isosceles right triangles with one<br />
leg in the disk. 8/3<br />
y<br />
√⎯⎯⎯⎯ sin x<br />
<br />
x<br />
43. Writing to Learn A solid lies between planes perpendicular<br />
to the x-axis at x 0 and x 12. The cross sections<br />
by planes perpendicular to the x-axis are circular disks whose<br />
diameters run from the line y x2 to the line y x as<br />
shown in the figure. Explain why the solid has the same volume<br />
as a right circular cone with base radius 3 and height 12.<br />
The volumes are<br />
y<br />
equal by Cavalieri’s<br />
Theorem.<br />
y x<br />
44. A Twisted Solid A square of side length s lies in a plane<br />
perpendicular to a line L. One vertex of the square lies on L.<br />
As this square moves a distance h along L, the square turns one<br />
revolution about L to generate a corkscrew-like column with<br />
square cross sections.<br />
(a) Find the volume of the column. s 2 h<br />
(b) Writing to Learn What will the volume be if the square<br />
turns twice instead of once? Give reasons for your answer. s 2 h<br />
45. Find the volume of the solid generated by revolving the region<br />
in the first quadrant bounded by y x 3 and y 4x about<br />
(a) the x-axis, 512p/21<br />
(b) the line y 8. 832p/21<br />
46. Find the volume of the solid generated by revolving the region<br />
bounded by y 2x x 2 and y x about<br />
(a) the y-axis, p/6<br />
(b) the line x 1. p/6<br />
47. The region in the first quadrant that is bounded above by the<br />
curve y 1x, on the left by the line x 14, and below<br />
by the line y 1 is revolved about the y-axis to generate a<br />
solid. Find the volume of the solid by (a) the washer method and<br />
(b) the cylindrical shell method. (a) 11p/48 (b) 11p/48<br />
sin xx, 0 x p<br />
48. Let f x { 1, x 0.<br />
0<br />
y 2<br />
x<br />
12<br />
x<br />
(a) Show that xfx sin x, 0 x p.<br />
(b) Find the volume of the solid generated by revolving the<br />
shaded region about the y-axis. 4p<br />
y<br />
1<br />
y<br />
y ⎧ sin<br />
——<br />
x<br />
,0 x ≤ <br />
⎨ x<br />
⎩ 1, x 0<br />
0<br />
x 2 y 2 1<br />
1<br />
x<br />
0 <br />
x
Section 7.3 Volumes 409<br />
49. Designing a Plumb Bob Having been asked to design<br />
a brass plumb bob that will weigh in the neighborhood of 190 g,<br />
you decide to shape it like the solid of revolution shown here.<br />
y (cm)<br />
y x 36 x 2<br />
12<br />
59. y x 2 , 0 x 2; x-axis 53.226<br />
60. y 3x x 2 , 0 x 3; x-axis 44.877<br />
61. y 2x x 2 , 0.5 x 1.5; x-axis 6.283<br />
62. y x 1, 1 x 5; x-axis 51.313<br />
0<br />
(a) Find the plumb bob’s volume. 36p/5 cm 3<br />
(b) If you specify a brass that weighs 8.5 gcm 3 , how much will<br />
the plumb bob weigh to the nearest gram? 192.3 g<br />
50. Volume of a Bowl A bowl has a shape that can be generated<br />
by revolving the graph of y x 2 2 between y 0 and y 5<br />
about the y-axis.<br />
(a) Find the volume of the bowl. 25p<br />
(b) If we fill the bowl with water at a constant rate of<br />
3 cubic units per second, how fast will the water level in<br />
the bowl be rising when the water is 4 units deep? 3/(8p)<br />
51. The Classical Bead Problem A round hole is drilled<br />
through the center of a spherical solid of radius r. The resulting<br />
cylindrical hole has height 4 cm.<br />
(a) What is the volume of the solid that remains? 32p/3<br />
(b) What is unusual about the answer? The answer is independent of r.<br />
52. Writing to Learn Explain how you could estimate the<br />
volume of a solid of revolution by measuring the shadow cast on<br />
a table parallel to its axis of revolution by a light shining directly<br />
above it. See page 410.<br />
53. Same Volume about Each Axis The region in the first<br />
quadrant enclosed between the graph of y ax x 2 and the<br />
x-axis generates the same volume whether it is revolved about<br />
the x-axis or the y-axis. Find the value of a. 5<br />
54. (Continuation of Exploration 2) Let x gy 0 have a<br />
continuous first derivative on c, d . Show that the area of the<br />
surface generated by revolving the curve x gy about the<br />
y-axis is See page 410.<br />
S d<br />
c<br />
2p gy 1 gy 2 dy.<br />
In Exercises 55–62, find the area of the surface generated by<br />
revolving the curve about the indicated axis.<br />
55. x y, 0 y 2; y-axis 13.614<br />
56. x y 3 3, 0 y 1; y-axis 0.638<br />
57. x y 12 13 32 , 1 y 3; y-axis 16.110<br />
58. x 2y 1, 58 y 1; y-axis 2.999<br />
6<br />
x (cm<br />
Standardized Test Questions<br />
You may use a graphing calculator to solve the following<br />
problems.<br />
63. True or False The volume of a solid of a known integrable<br />
cross section area A(x) from x a to x b is b A(x) dx. Justify<br />
a<br />
your answer. True, by definition.<br />
64. True or False If the region enclosed by the y-axis, the line<br />
y 2, and the curve y x is revolved about the y-axis, the<br />
volume of the solid is given by the definite integral 2 0 py2 dy.<br />
Justify your answer. False. The volume is given by 2<br />
py 4 dy.<br />
0<br />
65. Multiple Choice The base of a solid S is the region enclosed<br />
by the graph of y ln x, the line x e, and the x-axis. If the<br />
cross sections of S perpendicular to the x-axis are squares,<br />
which of the following gives the best approximation of the<br />
volume of S? A<br />
(A) 0.718 (B) 1.718 (C) 2.718 (D) 3.171 (E) 7.388<br />
66. Multiple Choice Let R be the region in the first quadrant<br />
bounded by the graph of y 8 x 3/2 , the x-axis, and the y-axis.<br />
Which of the following gives the best approximation of the<br />
volume of the solid generated when R is revolved about the<br />
x-axis? E<br />
(A) 60.3 (B) 115.2 (C) 225.4 (D) 319.7 (E) 361.9<br />
67. Multiple Choice Let R be the region enclosed by the graph of<br />
y x 2 , the line x 4, and the x-axis. Which of the following<br />
gives the best approximation of the volume of the solid generated<br />
when R is revolved about the y-axis? B<br />
(A) 64p (B) 128p (C) 256p (D) 360 (E) 512<br />
68. Multiple Choice Let R be the region enclosed by the graphs<br />
of y e x , y e x , and x 1. Which of the following gives the<br />
volume of the solid generated when R is revolved about the<br />
x-axis? D<br />
1<br />
(A) (e x e x ) dx<br />
0<br />
1<br />
(B) (e 2x e 2x ) dx<br />
0<br />
1<br />
(C) (e x e x ) 2 dx<br />
0<br />
1<br />
(D) p (e 2x e 2x ) dx<br />
0<br />
1<br />
(E) p (e x e x ) 2 dx<br />
0
410 Chapter 7 Applications of Definite Integrals<br />
Explorations<br />
69. Max-Min The arch y sin x, 0 x p, is revolved about the<br />
line y c, 0 c 1, to generate the solid in the figure.<br />
(a) Find the value of c that minimizes the volume of the solid.<br />
What is the minimum volume? p<br />
2 , <br />
p 2 2 8<br />
<br />
(b) What value of c in 0, 1 maximizes the volume of<br />
the solid? 0<br />
(c) Writing to Learn Graph the solid’s volume as a function<br />
of c, first for 0 c 1 and then on a larger domain. What<br />
happens to the volume of the solid as c moves away from 0, 1?<br />
Does this make sense physically? Give reasons for your answers.<br />
c<br />
y<br />
0<br />
y sin x<br />
y c<br />
70. A Vase We wish to estimate the volume of a flower vase using<br />
only a calculator, a string, and a ruler. We measure the height of the<br />
vase to be 6 inches. We then use the string and the ruler to find<br />
circumferences of the vase (in inches) at half-inch intervals. (We list<br />
them from the top down to correspond with the picture of the vase.)<br />
Circumferences<br />
6<br />
n 12. 34.7 in 3 5.4 10.8<br />
4.5 11.6<br />
4.4 11.6<br />
5.1 10.8<br />
6.3 9.0<br />
7.8 6.3<br />
0 9.4<br />
(a) Find the areas of the cross sections that correspond to the<br />
given circumferences. 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7,<br />
10.7, 9.3, 6.4, 3.2<br />
(b) Express the volume of the vase as an integral with respect to<br />
y over the interval 0, 6. 1<br />
4 p 6 C(y) 2 dy<br />
0<br />
(c) Approximate the integral using the Trapezoidal Rule with<br />
52. Partition the appropriate interval on the axis of revolution and measure the<br />
radius r(x) of the shadow region at these points. Then use an approximation<br />
such as the trapezoidal rule to estimate the integral b a pr2 (x) dx.<br />
54. For a tiny horizontal slice,<br />
slant height s (x) 2 (y) 2 1 (y)) (g 2 y. So the<br />
surface area is approximated by the Riemann sum<br />
n<br />
k1<br />
2p g(y k )1 (y)) (g 2 y.<br />
The limit of that is the integral.<br />
V p(2c2 p 8c p)<br />
<br />
2<br />
Volume →∞<br />
<br />
x<br />
Extending the Ideas<br />
71. Volume of a Hemisphere Derive the formula V 23 pR 3<br />
for the volume of a hemisphere of radius R by comparing its cross<br />
sections with the cross sections of a solid right circular cylinder of<br />
radius R and height R from which a solid right circular cone of base<br />
radius R and height R has been removed as suggested by the figure.<br />
h<br />
√⎺⎺⎺⎺⎺⎺ R 2 – h 2<br />
R<br />
72. Volume of a Torus The disk x 2 y 2 a 2 is revolved about<br />
the line x b b a to generate a solid shaped like a doughnut,<br />
called a torus. Find its volume. (Hint: a a a2 y 2 dy pa 2 2,<br />
since it is the area of a semicircle of radius a.) 2a 2 bp 2<br />
73. Filling a Bowl<br />
(a) Volume A hemispherical bowl of radius a contains water to<br />
a depth h. Find the volume of water in the bowl. ph 2 (3a h)/3<br />
(b) Related Rates Water runs into a sunken concrete<br />
hemispherical bowl of radius 5 m at a rate of 0.2 m 3 sec. How<br />
fast is the water level in the bowl rising when the water is 4 m<br />
deep? 1/(120p) m/sec<br />
74. Consistency of Volume Definitions The volume formulas<br />
in calculus are consistent with the standard formulas from<br />
geometry in the sense that they agree on objects to which both<br />
apply.<br />
(a) As a case in point, show that if you revolve the region<br />
enclosed by the semicircle y a 2 x <br />
2 and the x-axis about<br />
the x-axis to generate a solid sphere, the calculus formula for<br />
volume at the beginning of the section will give 43pa 3 for<br />
the volume just as it should.<br />
(b) Use calculus to find the volume of a right circular cone of<br />
height h and base radius r.<br />
71. Hemisphere cross sectional area:<br />
p(R 2 h<br />
2 ) 2 A 1<br />
Right circular cylinder with cone removed cross sectional area:<br />
pR 2 ph 2 A 2<br />
Since A 1 A 2 , the two volumes are equal by Cavalieri’s theorem.<br />
Thus,<br />
volume of hemisphere volume of cylinder volume of cone<br />
R<br />
h<br />
pR 3 1 3 pR 3 2 3 pR 3 .<br />
74. (a) A cross section has radius r a 2 x<br />
2 and<br />
area A(x) pr 2 p(a 2 x 2 ).<br />
V a p(a 2 x 2 ) 4<br />
a 3 pa3<br />
(b) A cross section has radius x r 1 y<br />
h and<br />
area A(y) px 2 pr 2 1 2 y<br />
y h h <br />
2 2 .<br />
V h<br />
pr 2<br />
0 1 2 y<br />
y 2<br />
<br />
h h <br />
dy 1 3 pr2 h<br />
h
Section 7.3 Volumes 411<br />
Quick Quiz for AP* Preparation: Sections 7.1–7.3<br />
You may use a graphing calculator to solve the following<br />
problems.<br />
1. Multiple Choice The base of a solid is the region in the first<br />
quadrant bounded by the x-axis, the graph of y sin 1 x, and the<br />
vertical line x 1. For this solid, each cross section perpendicular<br />
to the x-axis is a square. What is the volume? C<br />
(A) 0.117 (B) 0.285 (C) 0.467 (D) 0.571 (E) 1.571<br />
2. Multiple Choice Let R be the region in the first quadrant<br />
bounded by the graph of y 3x x 2 and the x-axis. A solid is<br />
generated when R is revolved about the vertical line x 1.<br />
Set up, but do not evaluate, the definite integral that gives the<br />
volume of this solid. A<br />
3<br />
(A) 2p(x 1)(3x x 2 ) dx<br />
(B)<br />
0<br />
3<br />
(C)<br />
1<br />
3<br />
0<br />
3<br />
(D)<br />
(E)<br />
0<br />
3<br />
0<br />
2p(x 1)(3x x 2 ) dx<br />
2p(x)(3x x 2 ) dx<br />
2p(3x x 2 ) 2 dx<br />
(3x x 2 ) dx<br />
3. Multiple Choice A developing country consumes oil at a rate<br />
given by r(t) 20e 0.2t million barrels per year, where t is time<br />
measured in years, for 0 t 10. Which of the following<br />
expressions gives the amount of oil consumed by the country<br />
during the time interval 0 t 10? D<br />
(A) r(10)<br />
(B) r(10) r(0)<br />
10<br />
(C) r(t) dt<br />
0<br />
10<br />
(D) r(t) dt<br />
0<br />
(E) 10 r(10)<br />
4. Free Response Let R be the region bounded by the graphs of<br />
y x, y e x , and the y-axis.<br />
(a) Find the area of R.<br />
(b) Find the volume of the solid generated when R is revolved<br />
about the horizontal line y 1.<br />
(c) The region R is the base of a solid. For this solid, each cross<br />
section perpendicular to the x-axis is a semicircle whose diameter<br />
runs from the graph of y x to the graph of y e x . Find<br />
the volume of this solid.<br />
4. (a) The two graphs intersect where x e –x , which a calculator shows to<br />
be x 0.42630275. Store this value as A.<br />
The area of R is A<br />
(e –x x) dx 0.162.<br />
0<br />
(b) Volume A<br />
p((e –x 1) 2 (x 1) 2 ) dx 1.631.<br />
0<br />
(c) Volume A<br />
1<br />
0 2 p x 2<br />
ex <br />
2 dx 0.035.
412 Chapter 7 Applications of Definite Integrals<br />
7.4<br />
What you’ll learn about<br />
• A Sine Wave<br />
• Length of a Smooth Curve<br />
• Vertical Tangents, Corners, and<br />
Cusps<br />
. . . and why<br />
The length of a smooth curve can<br />
be found using a definite integral.<br />
[0, 2] by [–2, 2]<br />
Figure 7.32 One wave of a sine curve<br />
has to be longer than 2p.<br />
y<br />
x y √<br />
2<br />
k + 2<br />
k<br />
Q<br />
y k<br />
y = sin x<br />
Lengths of Curves<br />
A Sine Wave<br />
How long is a sine wave (Figure 7.32)?<br />
The usual meaning of wavelength refers to the fundamental period, which for y sin x<br />
is 2p. But how long is the curve itself? If you straightened it out like a piece of string<br />
along the positive x-axis with one end at 0, where would the other end be?<br />
EXAMPLE 1<br />
The Length of a Sine Wave<br />
What is the length of the curve y sin x from x 0 to x 2p?<br />
SOLUTION<br />
We answer this question with integration, following our usual plan of breaking the whole<br />
into measurable parts. We partition 0, 2p into intervals so short that the pieces of curve<br />
(call them “arcs”) lying directly above the intervals are nearly straight. That way, each<br />
arc is nearly the same as the line segment joining its two ends and we can take the length<br />
of the segment as an approximation to the length of the arc.<br />
Figure 7.33 shows the segment approximating the arc above the subinterval x k1 , x k .<br />
The length of the segment is Δx 2 k . Δy 2 k The sum<br />
Δx k 2 Δy 2 k<br />
over the entire partition approximates the length of the curve. All we need now is to find<br />
the limit of this sum as the norms of the partitions go to zero. That’s the usual plan, but<br />
this time there is a problem. Do you see it?<br />
The problem is that the sums as written are not Riemann sums. They do not have the<br />
form f c k Δx. We can rewrite them as Riemann sums if we multiply and divide each<br />
square root by Δx k .<br />
O<br />
P<br />
x k–1<br />
x k<br />
Figure 7.33 The line segment approximating<br />
the arc PQ of the sine curve above<br />
the subinterval x k1 , x k . (Example 1)<br />
Group Exploration<br />
Later in this section we will use an integral<br />
to find the length of the sine wave<br />
with great precision. But there are ways<br />
to get good approximations without<br />
integrating. Take five minutes to come<br />
up with a written estimate of the curve’s<br />
length. No fair looking ahead.<br />
x k<br />
x<br />
x k<br />
2 y 2 k x k 2 y k<br />
<br />
2 x<br />
x<br />
k<br />
y<br />
x<br />
1 ( <br />
k<br />
)<br />
2<br />
k<br />
k<br />
This is better, but we still need to write the last square root as a function evaluated at<br />
some c k in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable<br />
functions (Section 4.2), which says that since sin x is continuous on x k1 , x k <br />
and is differentiable on x k1 , x k there is a point c k in x k1 , x k at which Δy k Δx k <br />
sin c k (Figure 7.34). That gives us<br />
1 si nc k <br />
2 Δx k ,<br />
which is a Riemann sum.<br />
Now we take the limit as the norms of the subdivisions go to zero and find that the<br />
length of one wave of the sine function is<br />
2p<br />
1 si nx 2 dx 2p<br />
1 cos 2 x dx 7.64. Using NINT<br />
0<br />
0<br />
How close was your estimate? Now try Exercise 9.<br />
x k
Section 7.4 Lengths of Curves 413<br />
P<br />
Slope sin' (c k )<br />
x k<br />
x k–1 c k x k<br />
y k<br />
Figure 7.34 The portion of the sine<br />
curve above x k1 , x k . At some c k in the<br />
interval, sin c k y k x k , the slope of<br />
segment PQ. (Example 1)<br />
d<br />
c<br />
0<br />
y<br />
(a, c)<br />
a<br />
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯<br />
(x k ) 2 (y k ) 2<br />
y f(x)<br />
x k – 1<br />
x k<br />
Q<br />
x k<br />
y k<br />
(b, d)<br />
Figure 7.35 The graph of f, approximated<br />
by line segments.<br />
P<br />
Q<br />
b<br />
x<br />
Length of a Smooth Curve<br />
We are almost ready to define the length of a curve as a definite integral, using the procedure<br />
of Example 1. We first call attention to two properties of the sine function that came<br />
into play along the way.<br />
We obviously used differentiability when we invoked the Mean Value Theorem to replace<br />
Δy k Δx k by sinc k for some c k in the interval x k1 , x k . Less obviously, we used<br />
the continuity of the derivative of sine in passing from 1 si nc k 2 Δx k to the<br />
Riemann integral. The requirement for finding the length of a curve by this method, then, is<br />
that the function have a continuous first derivative. We call this property smoothness. A<br />
function with a continuous first derivative is smooth and its graph is a smooth curve.<br />
Let us review the process, this time with a general smooth function f x. Suppose the<br />
graph of f begins at the point a, c and ends at b, d, as shown in Figure 7.35. We partition<br />
the interval a x b into subintervals so short that the arcs of the curve above them are<br />
nearly straight. The length of the segment approximating the arc above the subinterval<br />
x k1 , x k is Δx 2 k . Δy 2 k The sum Δx 2 k Δy 2 k approximates the length of the curve. We<br />
apply the Mean Value Theorem to f on each subinterval to rewrite the sum as a Riemann sum,<br />
x k<br />
2 y k<br />
y<br />
x<br />
2 1 ( <br />
)<br />
2<br />
k<br />
k<br />
x k<br />
1 fc k 2 x k .<br />
Passing to the limit as the norms of the subdivisions go to zero gives the length of the curve as<br />
L b<br />
a<br />
1 fx 2 dx b<br />
a<br />
1 ( d d<br />
We could as easily have transformed Δx 2 k Δy 2 k into a Riemann sum by dividing<br />
and multiplying by Δy k , giving a formula that involves x as a function of y say, x gy<br />
on the interval c, d:<br />
L x k 2 y 2 2<br />
k<br />
<br />
y<br />
y<br />
k ( 1 xk<br />
y<br />
y<br />
k<br />
k<br />
)<br />
k<br />
y<br />
) x<br />
2<br />
dx.<br />
1 gc 2 k y k .<br />
For some point<br />
c k in (x k 1 , x k )<br />
For some c k<br />
in (y k1 , y k )<br />
The limit of these sums, as the norms of the subdivisions go to zero, gives another reasonable<br />
way to calculate the curve’s length,<br />
L d<br />
1 gy 2 dy d<br />
2<br />
( 1 d x<br />
d<br />
) dy.<br />
y<br />
c<br />
c<br />
Putting these two formulas together, we have the following definition for the length of a<br />
smooth curve.<br />
DEFINITION<br />
Arc Length: Length of a Smooth Curve<br />
If a smooth curve begins at a, c and ends at b, d, a b, c d, then the length<br />
(arc length) of the curve is<br />
L b<br />
L d<br />
a<br />
1 ( d d<br />
c<br />
1 ( d d<br />
y<br />
) x<br />
2<br />
dx<br />
2<br />
x<br />
<br />
y ) dy<br />
if y is a smooth function of x on a, b;<br />
if x is a smooth function of y on c, d.
414 Chapter 7 Applications of Definite Integrals<br />
EXAMPLE 2<br />
Applying the Definition<br />
Find the exact length of the curve<br />
y 4 2<br />
x<br />
3<br />
1 for 0 x 1.<br />
SOLUTION<br />
d y<br />
4 2<br />
• 3 dx<br />
3 2 x12 22 x 12 ,<br />
which is continuous on 0, 1. Therefore,<br />
)<br />
L 1<br />
2<br />
( 1 d y<br />
dx<br />
dx<br />
0<br />
1<br />
( 22x12)<br />
2<br />
1 <br />
0<br />
1<br />
1 8x dx<br />
0<br />
2 1<br />
3 • 1 1 <br />
8<br />
8x32]<br />
0<br />
1 3<br />
.<br />
6<br />
dx<br />
y<br />
(8, 2)<br />
Now try Exercise 11.<br />
We asked for an exact length in Example 2 to take advantage of the rare opportunity it<br />
afforded of taking the antiderivative of an arc length integrand. When you add 1 to the<br />
square of the derivative of an arbitrary smooth function and then take the square root of that<br />
sum, the result is rarely antidifferentiable by reasonable methods. We know a few more<br />
functions that give “nice” integrands, but we are saving those for the exercises.<br />
(–8, –2)<br />
Figure 7.36 The graph of y x 13 has<br />
a vertical tangent line at the origin where<br />
dydx does not exist. (Example 3)<br />
(–2, –8)<br />
(2, 8)<br />
Figure 7.37 The curve in Figure 7.36<br />
plotted with x as a function of y. The<br />
tangent at the origin is now horizontal.<br />
(Example 3)<br />
x<br />
y<br />
x<br />
Vertical Tangents, Corners, and Cusps<br />
Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to<br />
work with is undefined. We can sometimes get around such a difficulty in ways illustrated by<br />
the following examples.<br />
EXAMPLE 3<br />
A Vertical Tangent<br />
Find the length of the curve y x 13 between 8, 2 and 8, 2.<br />
SOLUTION<br />
The derivative<br />
d y<br />
1 1<br />
dx<br />
3 x23 <br />
3x<br />
23<br />
is not defined at x 0. Graphically, there is a vertical tangent at x 0 where the derivative<br />
becomes infinite (Figure 7.36). If we change to x as a function of y, the tangent<br />
at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of<br />
undefined. Solving y x 13 for x gives x y 3 , and we have<br />
L 2<br />
2<br />
x<br />
<br />
y ) dy 2<br />
1 3y 2 <br />
2 dy 17.26. Using NINT<br />
2 1 ( d d<br />
2<br />
Now try Exercise 25.
Section 7.4 Lengths of Curves 415<br />
What happens if you fail to notice that dydx is undefined at x 0 and ask your calculator<br />
to compute<br />
2<br />
NINT<br />
(( 1 13 x23)<br />
8)<br />
, x, 8, ?<br />
This actually depends on your calculator. If, in the process of its calculations, it tries to<br />
evaluate the function at x 0, then some sort of domain error will result. If it tries to find<br />
convergent Riemann sums near x 0, it might get into a long, futile loop of computations<br />
that you will have to interrupt. Or it might actually produce an answer—in which<br />
case you hope it would be sufficiently bizarre for you to realize that it should not be<br />
trusted.<br />
EXAMPLE 4<br />
Getting Around a Corner<br />
Find the length of the curve y x 2 4x x from x 4 to x 4.<br />
SOLUTION<br />
We should always be alert for abrupt slope changes when absolute value is involved. We<br />
graph the function to check (Figure 7.38).<br />
There is clearly a corner at x 0 where neither dydx nor dxdy can exist. To find the<br />
length, we split the curve at x 0 to write the function without absolute values:<br />
[–5, 5] by [–7, 5]<br />
Figure 7.38 The graph of<br />
y x 2 4x x, 4 x 4,<br />
has a corner at x 0 where neither dy/dx<br />
nor dx/dy exists. We find the lengths of the<br />
two smooth pieces and add them together.<br />
(Example 4)<br />
Then,<br />
L 0<br />
4<br />
19.56.<br />
x 2 3x if x 0,<br />
x 2 4x x { x 2 5x if x 0.<br />
1 2x 3 2 dx 4<br />
1 2x 5 2 dx<br />
By NINT<br />
0<br />
Now try Exercise 27.<br />
Finally, cusps are handled the same way corners are: split the curve into smooth pieces<br />
and add the lengths of those pieces.<br />
Quick Review 7.4 (For help, go to Sections 1.3 and 3.2.)<br />
In Exercises 1–5, simplify the function.<br />
1. 1 2x x 2 on 1, 5 x 1<br />
2. <br />
1 <br />
x <br />
x 2<br />
4<br />
on 3, 1 2 2<br />
x<br />
<br />
3. 1 tan x 2 on 0, p3 sec x<br />
4. 1 x4 1x 2 on 4, 12 x2 4<br />
<br />
4x<br />
5. 1 cos2x on 0, p2 2 cos x<br />
In Exercises 6–10, identify all values of x for which the function fails<br />
to be differentiable.<br />
6. f x x 4 4<br />
7. f x 5x 23 0<br />
8. f x 5 x 3 3<br />
9. f x x 2 4x 4 2<br />
10. f x 1 3 si n x<br />
kp, k any integer
416 Chapter 7 Applications of Definite Integrals<br />
Section 7.4 Exercises<br />
In Exercises 1–10,<br />
(a) set up an integral for the length of the curve;<br />
(b) graph the curve to see what it looks like;<br />
(c) use NINT to find the length of the curve.<br />
1. y x 2 , 1 x 2<br />
2. y tan x, p3 x 0<br />
3. x sin y, 0 y p<br />
4. x 1 y , 2 12 y 12<br />
5. y 2 2y 2x 1, from 1, 1 to 7, 3<br />
6. y sin x x cos x, 0 x p<br />
x<br />
7. y 0 tan tdt, 0 x p6<br />
8. x y 0<br />
sec 2 t 1 dt, p3 y p4<br />
9. y sec x, p3 x p3<br />
10. y e x e x 2, 3 x 3<br />
In Exercises 11–18, find the exact length of the curve analytically by<br />
antidifferentiation. You will need to simplify the integrand<br />
algebraically before finding an antiderivative.<br />
11. y 13x 2 2 32 from x 0 to x 3 12<br />
12. y x 32 from x 0 to x 4 (8010 8)/27<br />
13. x y 3 3 14y from y 1 to y 3<br />
[Hint: 1 dxdy 2 is a perfect square.] 53/6<br />
14. x y 4 4 18y 2 from y 1 to y 2<br />
[Hint: 1 dxdy 2 is a perfect square.] 123/32<br />
15. x y 3 6 12y from y 1 to y 2<br />
[Hint: 1 dxdy 2 is a perfect square.] 17/12<br />
16. y x 3 3 x 2 x 14x 4, 0 x 2 53/6<br />
17. x y 0<br />
sec 4 t 1 dt, p4 y p4 2<br />
18. y x<br />
3t 4 1 dt, 2 x 1 73/3<br />
2<br />
19. (a) Group Activity Find a curve through the point 1, 1<br />
whose length integral is y x<br />
L 4<br />
1 1 4 x dx.<br />
1<br />
(b) Writing to Learn How many such curves are there? Give<br />
reasons for your answer.<br />
20. (a) Group Activity Find a curve through the point 0, 1<br />
whose length integral is y 1/(1 x)<br />
L 2<br />
1 y 1 4 dy.<br />
1<br />
(b) Writing to Learn How many such curves are there? Give<br />
reasons for your answer.<br />
21. Find the length of the curve<br />
y x<br />
from x 0 to x p4. 1<br />
0<br />
cos2t dt<br />
22. The Length of an Astroid The graph of the equation<br />
x 23 y 23 1 is one of the family of curves called astroids<br />
(not “asteroids”) because of their starlike appearance (see figure).<br />
Find the length of this particular astroid by finding the length of<br />
half the first quadrant portion, y 1 x 23 32 , 24 x 1,<br />
and multiplying by 8. 6<br />
y<br />
23. Fabricating Metal Sheets Your metal fabrication company is<br />
bidding for a contract to make sheets of corrugated steel roofing<br />
like the one shown here. The cross sections of the corrugated<br />
sheets are to conform to the curve<br />
y sin ( 3 p<br />
20 ) x , 0 x 20 in.<br />
If the roofing is to be stamped from flat sheets by a process that<br />
does not stretch the material, how wide should the original<br />
material be? Give your answer to two decimal places. 21.07 inches<br />
Original sheet<br />
–1<br />
20 in.<br />
Corrugated sheet<br />
24. Tunnel Construction Your engineering firm is bidding for<br />
the contract to construct the tunnel shown on the next page. The<br />
tunnel is 300 ft long and 50 ft wide at the base. The cross<br />
section is shaped like one arch of the curve y 25 cos px50.<br />
Upon completion, the tunnel’s inside surface (excluding the<br />
roadway) will be treated with a waterproof sealer that costs<br />
$1.75 per square foot to apply. How much will it cost to apply<br />
the sealer? $38,422<br />
1<br />
0<br />
–1<br />
O<br />
y<br />
x 2/3 y 2/3 1<br />
y sin —–<br />
3<br />
x<br />
20<br />
20<br />
1<br />
x<br />
x (in.)<br />
19. (b) Only one. We know the derivative of the function and the value of the<br />
function at one value of x.<br />
20. (b) Only one. We know the derivative of the function and the value of the<br />
function at one value of x.
30. Because the limit of the sum x k as the norm of the partition goes to<br />
zero will always be the length (b a) of the interval (a, b).<br />
y<br />
y 25 cos (x/50)<br />
Section 7.4 Lengths of Curves 417<br />
34. Multiple Choice Which of the following gives the best<br />
approximation of the length of the arc of y cos(2x) from x 0<br />
to x p/4? D<br />
(A) 0.785 (B) 0.955 (C) 1.0 (D) 1.318 (E) 1.977<br />
–25<br />
0<br />
25<br />
NOT TO SCALE<br />
300 ft<br />
x (ft)<br />
35. Multiple Choice Which of the following expressions gives<br />
the length of the graph of x y 3 from y 2 to y 2? C<br />
2<br />
(A)2<br />
2<br />
(1 y 6 ) dy (B)<br />
2<br />
(C) 1 9y 4 dy<br />
2<br />
2<br />
(E) 1 x 4 dx<br />
2<br />
2<br />
1 y 6 dy<br />
2<br />
(D) 1 x 2 dx<br />
2<br />
36. Multiple Choice Find the length of the curve described by<br />
y 2 3 x3/2 from x 0 to x 8.<br />
B<br />
(A) 2 6<br />
(B) 5 2<br />
(C) 512 2<br />
<br />
3<br />
3<br />
15<br />
(D) 512 2<br />
8 (E) 96<br />
15<br />
In Exercises 25 and 26, find the length of the curve.<br />
25. f x x 13 x 23 , 0 x 2 3.6142<br />
x 1<br />
26. f x 4 x<br />
<br />
2 , 1 1 2 x 1 2.1089<br />
In Exercises 27–29, find the length of the nonsmooth curve.<br />
27. y x 3 5x from x 2 to x 1 13.132<br />
28. x y 1 1.623<br />
29. y 4 x from x 0 to x 16 16.647<br />
30. Writing to Learn Explain geometrically why it does not work<br />
to use short horizontal line segments to approximate the lengths<br />
of small arcs when we search for a Riemann sum that leads to the<br />
formula for arc length.<br />
31. Writing to Learn A curve is totally contained inside the<br />
square with vertices 0, 0, 1, 0, 1, 1, and 0, 1. Is there any<br />
limit to the possible length of the curve? Explain.<br />
Standardized Test Questions<br />
You should solve the following problems without using<br />
a graphing calculator.<br />
32. True or False If a function y f (x) is continuous on an<br />
interval [a, b], then the length of its curve is given by<br />
b<br />
1 d y a<br />
d<br />
x 2<br />
dx. Justify your answer.<br />
False. The function must be differentiable.<br />
33. True or False If a function y f (x) is differentiable on an<br />
interval [a, b], then the length of its curve is given by<br />
b<br />
y x 2<br />
dx. Justify your answer.<br />
True, by definition.<br />
a<br />
1 d d<br />
31. No. Consider the curve y 1 3 sin 1 x <br />
0.5 for 0 x 1.<br />
37. Multiple Choice Which of the following expressions should<br />
be used to find the length of the curve y x 2/3 from x 1<br />
to x 1? A<br />
1<br />
(A) 2 1 9 0<br />
4 y<br />
<br />
1<br />
(C)0<br />
1<br />
(E)<br />
0<br />
1 y 3 dy<br />
1 y 9/4<br />
dy<br />
Exploration<br />
1<br />
dy (B)1 1 9 4 y<br />
<br />
dy<br />
1<br />
(D) 1 y 6 dy<br />
0<br />
38. Modeling Running Tracks Two lanes of a running track<br />
are modeled by the semiellipses as shown. The equation for<br />
lane 1 is y 100 0.2x 2 , and the equation for lane 2<br />
is y 15 0 0.2x 2 . The starting point for lane 1 is at the<br />
negative x-intercept 50 0,0. The finish points for both lanes<br />
are the positive x-intercepts. Where should the starting point be<br />
placed on lane 2 so that the two lane lengths will be equal<br />
(running clockwise)? (–19.909, 8.410)<br />
Start lane 2<br />
Start lane 1<br />
1<br />
y<br />
10<br />
x
418 Chapter 7 Applications of Definite Integrals<br />
Extending the Ideas<br />
39. Using Tangent Fins to Find Arc Length Assume f is<br />
smooth on a, b and partition the interval a, b in the usual<br />
way. In each subinterval x k1 , x k construct the tangent fin at<br />
the point x k1 , f x k1 as shown in the figure.<br />
(x k –1 , f(x k –1 ))<br />
y f(x)<br />
x k<br />
Tangent fin<br />
with slope<br />
f'(x k –1 )<br />
(a) Show that the length of the kth tangent fin over the interval<br />
x k1 , x k equals<br />
x 2<br />
k f x k1 x 2 k .<br />
(b) Show that<br />
lim<br />
n→∞<br />
n<br />
(length of kth tangent fin) b<br />
1 fx 2 dx,<br />
k1<br />
a<br />
which is the length L of the curve y f x from x a<br />
to x b.<br />
40. Is there a smooth curve y f x whose length over<br />
the interval 0 x a is always a2? Give reasons<br />
for your answer. Yes. Any curve of the form y x c, c a<br />
constant.<br />
x k –1<br />
x k<br />
x<br />
39. (a) The fin is the hypotenuse of a right triangle with leg lengths x k and<br />
df<br />
d x<br />
xxk1<br />
x k f(x k–1 ) x k .<br />
(b) lim<br />
n→∞<br />
n<br />
( x k ) 2 ( f(x x k1 )<br />
k ) 2<br />
k1<br />
lim<br />
n→∞<br />
n<br />
x k 1(x ( f) k1 ) <br />
2<br />
k1<br />
1(x)) ( f 2 dx<br />
b<br />
a
Section 7.5 Applications from Science and Statistics 419<br />
7.5<br />
What you’ll learn about<br />
• Work Revisited<br />
• Fluid Force and Fluid Pressure<br />
• Normal Probabilities<br />
. . . and why<br />
It is important to see applications<br />
of integrals as various accumulation<br />
functions.<br />
Applications from Science and Statistics<br />
Our goal in this section is to hint at the diversity of ways in which the definite integral can<br />
be used. The contexts may be new to you, but we will explain what you need to know as we<br />
go along.<br />
Work Revisited<br />
Recall from Section 7.1 that work is defined as force (in the direction of motion) times displacement.<br />
A familiar example is to move against the force of gravity to lift an object. The<br />
object has to move, incidentally, before “work” is done, no matter how tired you get trying.<br />
If the force F(x) is not constant, then the work done in moving an object from x a to<br />
x b is the definite integral W b a F(x)dx.<br />
4.4 newtons 1 lb<br />
(1 newton)(1 meter) 1 N•m 1 Joule<br />
EXAMPLE 1<br />
Finding the Work Done by a Force<br />
Find the work done by the force F(x) cos(px) newtons along the x-axis from x 0<br />
meters to x 12 meter.<br />
SOLUTION<br />
12<br />
W cos(px) dx<br />
0<br />
1 1/2<br />
sin(px) p 0<br />
p<br />
1 sin p 2 sin(0) <br />
p<br />
1 0.318<br />
Now try Exercise 1.<br />
22<br />
(N)<br />
Work<br />
440<br />
N. m<br />
EXAMPLE 2<br />
Work Done Lifting<br />
A leaky bucket weighs 22 newtons (N) empty. It is lifted from the ground at a constant<br />
rate to a point 20 m above the ground by a rope weighing 0.4 N/m. The bucket starts<br />
with 70 N (approximately 7.1 liters) of water, but it leaks at a constant rate and just<br />
finishes draining as the bucket reaches the top. Find the amount of work done<br />
(a) lifting the bucket alone;<br />
(b) lifting the water alone;<br />
(c) lifting the rope alone;<br />
(d) lifting the bucket, water, and rope together.<br />
SOLUTION<br />
(a) The bucket alone. This is easy because the bucket’s weight is constant. To lift it, you<br />
must exert a force of 22 N through the entire 20-meter interval.<br />
(m) 20<br />
Figure 7.39 The work done by a<br />
constant 22-N force lifting a bucket 20 m<br />
is 440 N • m. (Example 2)<br />
Work 22 N 20 m 440 N • m 440 J<br />
Figure 7.39 shows the graph of force vs. distance applied. The work corresponds to the<br />
area under the force graph.<br />
continued
420 Chapter 7 Applications of Definite Integrals<br />
70<br />
(N)<br />
y<br />
F(x) = 70 – 3.5x<br />
(m) 20<br />
Figure 7.40 The force required to lift<br />
the water varies with distance but the work<br />
still corresponds to the area under the force<br />
graph. (Example 2)<br />
8<br />
(N)<br />
Work<br />
(m)<br />
20<br />
Figure 7.41 The work done lifting the<br />
rope to the top corresponds to the area of<br />
another triangle. (Example 2)<br />
x<br />
(b) The water alone. The force needed to lift the water is equal to the water’s weight,<br />
which decreases steadily from 70 N to 0 N over the 20-m lift. When the bucket is x m off<br />
the ground, the water weighs<br />
Fx 70 ( 20 x<br />
20<br />
) ( 70 1 x<br />
70 3.5x N.<br />
20<br />
The work done is (Figure 7.40)<br />
W b<br />
a<br />
20<br />
0<br />
Fx dx<br />
70 3.5x dx <br />
[<br />
)<br />
20<br />
70x 1.75x2]<br />
1400 700 700 J.<br />
0<br />
(c) The rope alone. The force needed to lift the rope is also variable, starting at<br />
0.420 8 N when the bucket is on the ground and ending at 0 N when the bucket<br />
and rope are all at the top. As with the leaky bucket, the rate of decrease is constant.<br />
At elevation x meters, the 20 x meters of rope still there to lift weigh Fx 0.4<br />
20 x N. Figure 7.41 shows the graph of F. The work done lifting the rope is<br />
Fx dx 20<br />
0.420 x dx<br />
20<br />
0<br />
original weight<br />
of water<br />
proportion left<br />
at elevation x<br />
0<br />
20<br />
<br />
[<br />
8x 0.2x2]<br />
160 80 80 N • m 80 J.<br />
0<br />
(d) The bucket, water, and rope together. The total work is<br />
440 700 80 1220 J. Now try Exercise 5.<br />
10 y<br />
y<br />
10<br />
8<br />
0<br />
y<br />
Δy<br />
y 2x or x<br />
1<br />
y<br />
2<br />
1<br />
y<br />
2<br />
5<br />
(5, 10)<br />
Figure 7.42 The conical tank in<br />
Example 3.<br />
x<br />
EXAMPLE 3<br />
Work Done Pumping<br />
The conical tank in Figure 7.42 is filled to within 2 ft of the top with olive oil weighing<br />
57 lbft 3 . How much work does it take to pump the oil to the rim of the tank?<br />
SOLUTION<br />
We imagine the oil partitioned into thin slabs by planes perpendicular to the y-axis<br />
at the points of a partition of the interval 0, 8. (The 8 represents the top of the oil,<br />
not the top of the tank.)<br />
The typical slab between the planes at y and y Δy has a volume of about<br />
2<br />
Δy p 4 y2 Δy ft 3 .<br />
ΔV p(radius) 2 (thickness) p ( 1 2 y )<br />
The force Fy required to lift this slab is equal to its weight,<br />
Fy 57 ΔV 57 p<br />
y<br />
4<br />
2 Δy lb.<br />
weight per<br />
Weight ( unit volume ) volume<br />
The distance through which Fy must act to lift this slab to the level of the rim of the<br />
cone is about 10 y ft, so the work done lifting the slab is about<br />
ΔW 57 p<br />
10 yy<br />
4<br />
2 Δy ft • lb.<br />
The work done lifting all the slabs from y 0 to y 8 to the rim is approximately<br />
W 57 p<br />
10 y y<br />
4<br />
2 Δy ft • lb.<br />
continued
Section 7.5 Applications from Science and Statistics 421<br />
This is a Riemann sum for the function 57p410 yy 2 on the interval from y 0<br />
to y 8. The work of pumping the oil to the rim is the limit of these sums as the<br />
norms of the partitions go to zero.<br />
57 p<br />
10 yy<br />
4<br />
2 dy 57 p<br />
10y<br />
4<br />
2 y 3 dy<br />
0<br />
0<br />
57 p<br />
4<br />
[ 10 y<br />
3<br />
y 4<br />
8<br />
3 4<br />
]<br />
30,561 ft • lb<br />
0<br />
Now try Exercise 17.<br />
W 8<br />
8<br />
Figure 7.43 To withstand the increasing<br />
pressure, dams are built thicker toward the<br />
bottom.<br />
Fluid Force and Fluid Pressure<br />
We make dams thicker at the bottom than at the top (Figure 7.43) because the pressure<br />
against them increases with depth. It is a remarkable fact that the pressure at any point on<br />
a dam depends only on how far below the surface the point lies and not on how much<br />
water the dam is holding back. In any liquid, the fluid pressure p (force per unit area) at<br />
depth h is<br />
lb<br />
b<br />
p wh, Dimensions check: f t2 f<br />
lt3 ft, for example<br />
where w is the weight-density (weight per unit volume) of the liquid.<br />
EXAMPLE 4 The Great Molasses Flood of 1919<br />
Typical Weight-densities (lb/ft 3 )<br />
Gasoline 42<br />
Mercury 849<br />
Milk 64.5<br />
Molasses 100<br />
Seawater 64<br />
Water 62.4<br />
90 ft<br />
1 ft<br />
SHADED BAND NOT TO SCALE<br />
90 ft<br />
Figure 7.44 The molasses tank of<br />
Example 4.<br />
At 1:00 P.M. on January 15, 1919 (an unseasonably warm day), a 90-ft-high, 90-footdiameter<br />
cylindrical metal tank in which the Puritan Distilling Company stored molasses at<br />
the corner of Foster and Commercial streets in Boston’s North End exploded. Molasses<br />
flooded the streets 30 feet deep, trapping pedestrians and horses, knocking down buildings,<br />
and oozing into homes. It was eventually tracked all over town and even made its<br />
way into the suburbs via trolley cars and people’s shoes. It took weeks to clean up.<br />
(a) Given that the tank was full of molasses weighing 100 lbft 3 , what was the total<br />
force exerted by the molasses on the bottom of the tank at the time it ruptured?<br />
(b) What was the total force against the bottom foot-wide band of the tank wall<br />
(Figure 7.44)?<br />
continued
422 Chapter 7 Applications of Definite Integrals<br />
SOLUTION<br />
(a) At the bottom of the tank, the molasses exerted a constant pressure of<br />
90 y k<br />
45<br />
Figure 7.45 The 1-ft band at the bottom<br />
of the tank wall can be partitioned into thin<br />
strips on which the pressure is approximately<br />
constant. (Example 4)<br />
p wh ( 100 lb<br />
ft3) (90 ft) 9000 lb<br />
ft2.<br />
Since the area of the base was p45 2 , the total force on the base was<br />
(<br />
lb<br />
9000 f t2) (2025 p ft2 ) 57,225,526 lb.<br />
(b) We partition the band from depth 89 ft to depth 90 ft into narrower bands of width<br />
Δy and choose a depth y k in each one. The pressure at this depth y k is p wh 100 y k<br />
lbft 2 (Figure 7.45). The force against each narrow band is approximately<br />
pressure area 100y k 90p Δy 9000p y k Δy lb.<br />
Adding the forces against all the bands in the partition and passing to the limit as the<br />
norms go to zero, we arrive at<br />
F 90<br />
89<br />
9000pydy 9000p 90<br />
ydy 2,530,553 lb<br />
89<br />
for the force against the bottom foot of tank wall. Now try Exercise 25.<br />
1<br />
12<br />
y<br />
2<br />
Area<br />
= 1<br />
4<br />
5<br />
12<br />
Figure 7.46 The probability that the<br />
clock has stopped between 2:00 and 5:00<br />
can be represented as an area of 14.<br />
The rectangle over the entire interval has<br />
area 1.<br />
x<br />
Normal Probabilities<br />
Suppose you find an old clock in the attic. What is the probability that it has stopped<br />
somewhere between 2:00 and 5:00?<br />
If you imagine time being measured continuously over a 12-hour interval, it is easy to<br />
conclude that the answer is 14 (since the interval from 2:00 to 5:00 contains one-fourth of<br />
the time), and that is correct. Mathematically, however, the situation is not quite that clear<br />
because both the 12-hour interval and the 3-hour interval contain an infinite number of<br />
times. In what sense does the ratio of one infinity to another infinity equal 14?<br />
The easiest way to resolve that question is to look at area. We represent the total probability<br />
of the 12-hour interval as a rectangle of area 1 sitting above the interval (Figure 7.46).<br />
Not only does it make perfect sense to say that the rectangle over the time interval 2, 5<br />
has an area that is one-fourth the area of the total rectangle, the area actually equals 14,<br />
since the total rectangle has area 1. That is why mathematicians represent probabilities as<br />
areas, and that is where definite integrals enter the picture.<br />
Improper Integrals<br />
More information about improper<br />
integrals like <br />
fx dx can be found in<br />
<br />
Section 8.3. (You will not need that<br />
information here.)<br />
DEFINITION Probability Density Function (pdf)<br />
A probability density function is a function f x with domain all reals such that<br />
<br />
f x 0 for all x and f x dx 1. <br />
Then the probability associated with an interval a, b is<br />
b<br />
f x dx.<br />
a<br />
Probabilities of events, such as the clock stopping between 2:00 and 5:00, are integrals<br />
of an appropriate pdf.
Section 7.5 Applications from Science and Statistics 423<br />
EXAMPLE 5<br />
Probability of the Clock Stopping<br />
Find the probability that the clock stopped between 2:00 and 5:00.<br />
SOLUTION<br />
The pdf of the clock is<br />
112, 0 t 12<br />
f t { 0, otherwise.<br />
The probability that the clock stopped at some time t with 2 t 5 is<br />
5<br />
f t dt 1 4 . 2<br />
Now try Exercise 27.<br />
∫f(x)dx = .17287148<br />
Figure 7.47 A normal probability density<br />
function. The probability associated with<br />
the interval a, b is the area under the<br />
curve, as shown.<br />
a<br />
b<br />
By far the most useful kind of pdf is the normal kind. (“Normal” here is a technical<br />
term, referring to a curve with the shape in Figure 7.47.) The normal curve, often called<br />
the “bell curve,” is one of the most significant curves in applied mathematics because it<br />
enables us to describe entire populations based on the statistical measurements taken from a<br />
reasonably-sized sample. The measurements needed are the mean m and the standard<br />
deviation s, which your calculators will approximate for you from the data. The symbols<br />
on the calculator will probably be Jx and s (see your Owner’s Manual), but go ahead and use<br />
them as m and s, respectively. Once you have the numbers, you can find the curve by using<br />
the following remarkable formula discovered by Karl Friedrich Gauss.<br />
DEFINITION<br />
Normal Probability Density Function (pdf)<br />
The normal probability density function (Gaussian curve) for a population with<br />
mean m and standard deviation s is<br />
1<br />
f x e xm2 2s 2 .<br />
s 2p<br />
68% of area<br />
–3<br />
95% of the area<br />
99.7% of the area<br />
–2 –1 0 1 2 3<br />
Figure 7.48 The 68-95-99.7 rule for<br />
normal distributions.<br />
The mean m represents the average value of the variable x. The standard deviation s<br />
measures the “scatter” around the mean. For a normal curve, the mean and standard deviation<br />
tell you where most of the probability lies. The rule of thumb, illustrated in Figure 7.48,<br />
is this:<br />
The 68-95-99.7 Rule for Normal Distributions<br />
y<br />
0 <br />
= 0.5<br />
= 1<br />
= 2<br />
Figure 7.49 Normal pdf curves with<br />
mean m 2 and s 0.5, 1, and 2.<br />
x<br />
Given a normal curve,<br />
• 68% of the area will lie within s of the mean m,<br />
• 95% of the area will lie within 2s of the mean m,<br />
• 99.7% of the area will lie within 3s of the mean m.<br />
Even with the 68-95-99.7 rule, the area under the curve can spread quite a bit, depending<br />
on the size of s. Figure 7.49 shows three normal pdfs with mean m 2 and standard<br />
deviations equal to 0.5, 1, and 2.
424 Chapter 7 Applications of Definite Integrals<br />
EXAMPLE 6<br />
A Telephone Help Line<br />
Suppose a telephone help line takes a mean of 2 minutes to answer calls. If the standard<br />
deviation is s 0.5, then 68% of the calls are answered in the range of 1.5 to 2.5 minutes<br />
and 99.7% of the calls are answered in the range of 0.5 to 3.5 minutes.<br />
Now try Exercise 29.<br />
10.3 11<br />
[9, 11.5] by [–1, 2.5]<br />
Figure 7.50 The normal pdf for the<br />
spinach weights in Example 7. The mean<br />
is at the center.<br />
EXAMPLE 7<br />
Weights of Spinach Boxes<br />
Suppose that frozen spinach boxes marked as “10 ounces” of spinach have a mean<br />
weight of 10.3 ounces and a standard deviation of 0.2 ounce.<br />
(a) What percentage of all such spinach boxes can be expected to weigh between 10 and<br />
11 ounces?<br />
(b) What percentage would we expect to weigh less than 10 ounces?<br />
(c) What is the probability that a box weighs exactly 10 ounces?<br />
SOLUTION<br />
Assuming that some person or machine is trying to pack 10 ounces of spinach into these<br />
boxes, we expect that most of the weights will be around 10, with probabilities tailing<br />
off for boxes being heavier or lighter. We expect, in other words, that a normal pdf will<br />
model these probabilities. First, we define f x using the formula:<br />
1<br />
f x e x10.320.08 .<br />
0.2 2p<br />
The graph (Figure 7.50) has the look we are expecting.<br />
(a) For an arbitrary box of this spinach, the probability that it weighs between 10 and 11<br />
ounces is the area under the curve from 10 to 11, which is<br />
NINT f x, x, 10, 11 0.933.<br />
So without doing any more measuring, we can predict that about 93.3% of all such<br />
spinach boxes will weigh between 10 and 11 ounces.<br />
(b) For the probability that a box weighs less than 10 ounces, we use the entire area<br />
under the curve to the left of x 10. The curve actually approaches the x-axis as an<br />
asymptote, but you can see from the graph (Figure 7.50) that f x approaches zero quite<br />
quickly. Indeed, f 9 is only slightly larger than a billionth. So getting the area from 9 to<br />
10 should do it:<br />
NINT f x, x, 9,100.067.<br />
We would expect only about 6.7% of the boxes to weigh less than 10 ounces.<br />
(c) This would be the integral from 10 to 10, which is zero. This zero probability might<br />
seem strange at first, but remember that we are assuming a continuous, unbroken interval<br />
of possible spinach weights, and 10 is but one of an infinite number of them.<br />
Now try Exercise 31.<br />
Quick Review 7.5 (For help, go to Section 5.2.)<br />
In Exercises 1–5, find the definite integral by (a) antiderivatives and<br />
(b) using NINT.<br />
1. 1<br />
e x dx 2. 1<br />
e x dx a. e 1 b. 1.718<br />
a. 1 (1/e) b. 0.632<br />
0<br />
0<br />
p2 a. 2/2 b. 0.707<br />
3. sin xdx 4. 3<br />
x 2 2 dx 15<br />
p4<br />
0<br />
5. 2<br />
x2<br />
x<br />
3<br />
dx a. (1/3) ln (9/2) b. 0.501<br />
1<br />
1
Section 7.5 Applications from Science and Statistics 425<br />
In Exercises 6–10 find, but do not evaluate, the definite integral that<br />
is the limit as the norms of the partitions go to zero of the Riemann<br />
sums on the closed interval 0, 7.<br />
6. 2px k 2sin x k Δ x 7<br />
7. 1 x k 2 2p x k Δx 7<br />
0<br />
0<br />
2p(x 2) sin x dx<br />
(1 x 2 )(2px) dx<br />
8. pcos x k 2 Δx 7<br />
9. <br />
k)2<br />
p( y 2<br />
3<br />
4<br />
0<br />
p cos 2 x dx<br />
10 y k Δy 7<br />
p(y/2) 2 (10 y) dy<br />
10. sin 2 x k Δ x 7<br />
0<br />
0<br />
(3/4) sin 2 x dx<br />
Section 7.5 Exercises<br />
In Exercises 1–4, find the work done by the force of F(x) newtons<br />
along the x-axis from x a meters to x b meters.<br />
1. Fx xe x3 , a 0, b 5 4.4670 J<br />
2. Fx x sin px4, a 0, b 3 3.8473 J<br />
3. Fx x9 x 2 , a 0, b 3 9 J<br />
4. Fx e sin x cos x 2, a 0, b 10 19.5804 J<br />
5. Leaky Bucket The workers in Example 2 changed to a larger<br />
bucket that held 50 L (490 N) of water, but the new bucket had<br />
an even larger leak so that it too was empty by the time it<br />
reached the top. Assuming the water leaked out at a steady rate,<br />
how much work was done lifting the water to a point 20 meters<br />
above the ground? (Do not include the rope and bucket.) 4900 J<br />
6. Leaky Bucket The bucket in Exercise 5 is hauled up more<br />
quickly so that there is still 10 L (98 N) of water left when the<br />
bucket reaches the top. How much work is done lifting the water<br />
this time? (Do not include the rope and bucket.) 5880 J<br />
7. Leaky Sand Bag A bag of sand originally weighing 144 lb<br />
was lifted at a constant rate. As it rose, sand leaked out at a<br />
constant rate. The sand was half gone by the time the bag had<br />
been lifted 18 ft. How much work was done lifting the sand this<br />
far? (Neglect the weights of the bag and lifting equipment.) 1944 ft-lb<br />
8. Stretching a Spring A spring has a natural length of 10 in.<br />
An 800-lb force stretches the spring to 14 in.<br />
(a) Find the force constant. 200 lb/in.<br />
(b) How much work is done in stretching the spring from 10 in.<br />
to 12 in.? 400 in.-lb<br />
(c) How far beyond its natural length will a 1600-lb force<br />
stretch the spring? 8 in.<br />
9. Subway Car Springs It takes a force of 21,714 lb to compress<br />
a coil spring assembly on a New York City Transit<br />
Authority subway car from its free height of 8 in. to its fully<br />
compressed height of 5 in.<br />
(a) What is the assembly’s force constant? 7238 lb/in.<br />
(b) How much work does it take to compress the assembly the<br />
first half inch? the second half inch? Answer to the nearest inchpound.<br />
905 in.-lb and 2714 in.-lb<br />
(Source: Data courtesy of Bombardier, Inc., Mass Transit<br />
Division, for spring assemblies in subway cars delivered to the<br />
New York City Transit Authority from 1985 to 1987.)<br />
10. Bathroom Scale A bathroom scale is compressed 116 in.<br />
when a 150-lb person stands on it. Assuming the scale behaves<br />
like a spring that obeys Hooke’s Law,<br />
(a) how much does someone who compresses the scale<br />
18 in. weigh? 300 lb<br />
18.75 in.-lb<br />
(b) how much work is done in compressing the scale 18 in.?<br />
11. Hauling a Rope A mountain climber is about to haul up a<br />
50-m length of hanging rope. How much work will it take if the<br />
rope weighs 0.624 Nm? 780 J<br />
12. Compressing Gas Suppose that gas in a circular cylinder<br />
of cross section area A is being compressed by a piston<br />
(see figure).<br />
y<br />
(a) If p is the pressure of the gas in pounds per square inch and<br />
V is the volume in cubic inches, show that the work done in<br />
compressing the gas from state p 1 , V 1 to state p 2 , V 2 is<br />
given by the equation<br />
p 2 , V 2 <br />
Work pdV in. • lb,<br />
p 1 , V 1 <br />
where the force against the piston is pA.<br />
(b) Find the work done in compressing the gas from<br />
V 1 243 in 3 to V 2 32 in 3 if p 1 50 lbin 3 and<br />
p and V obey the gas law pV 1.4 constant (for adiabatic<br />
processes). –37,968.75 in.-lb<br />
x
426 Chapter 7 Applications of Definite Integrals<br />
Group Activity In Exercises 13–16, the vertical end of a tank<br />
containing water (blue shading) weighing 62.4 lbft 3 has the given<br />
shape.<br />
(a) Writing to Learn Explain how to approximate the force<br />
against the end of the tank by a Riemann sum.<br />
(b) Find the force as an integral and evaluate it.<br />
13. semicircle (b) 1123.2 lb 14. semiellipse (b) 7987.2 lb<br />
15. triangle (b) 3705 lb 16. parabola (b) 1506.1 lb<br />
3<br />
3 ft<br />
8 ft<br />
6 ft<br />
6 ft<br />
8 ft<br />
4 ft 4.5 ft<br />
17. (d) 1,494,240 ft-lb, 100 min; 1,500,000 ft-lb, 100 min<br />
(d) The Weight of Water Because of differences in the<br />
strength of Earth’s gravitational field, the weight of a cubic foot<br />
of water at sea level can vary from as little as 62.26 lb at the<br />
equator to as much as 62.59 lb near the poles, a variation of<br />
about 0.5%. A cubic foot of water that weighs 62.4 lb in<br />
Melbourne or New York City will weigh 62.5 lb in Juneau or<br />
Stockholm. What are the answers to parts (a) and (b) in a<br />
location where water weighs 62.26 lbft 3 ? 62.5 lbft 3 ?<br />
18. Emptying a Tank A vertical right cylindrical tank measures<br />
30 ft high and 20 ft in diameter. It is full of kerosene weighing<br />
51.2 lbft 3 . How much work does it take to pump the kerosene<br />
to the level of the top of the tank? 7,238,229 ft-lb<br />
19. Writing to Learn The cylindrical tank shown here is to be<br />
filled by pumping water from a lake 15 ft below the bottom<br />
of the tank. There are two ways to go about this. One is to pump<br />
the water through a hose attached to a valve in the bottom of the<br />
tank. The other is to attach the hose to the rim of the tank and let<br />
the water pour in. Which way will require less work? Give<br />
reasons for your answer.<br />
Through valve:<br />
84,687.3 ft-lb<br />
Over the rim:<br />
98,801.8 ft-lb<br />
Through a hose attached<br />
to a valve in the bottom is<br />
faster, because it takes<br />
more time to do more<br />
work.<br />
2 ft<br />
Open top<br />
6 ft<br />
6 ft<br />
17. Pumping Water The rectangular tank shown here, with its top<br />
at ground level, is used to catch runoff water. Assume that the<br />
water weighs 62.4 lbft 3 .<br />
Ground<br />
level<br />
10 ft<br />
0 12 ft<br />
Valve at base<br />
20. Drinking a Milkshake The truncated conical container shown<br />
here is full of strawberry milkshake that weighs 49 ozin 3 .<br />
As you can see, the container is 7 in. deep, 2.5 in. across at the<br />
base, and 3.5 in. across at the top (a standard size at Brigham’s<br />
in Boston). The straw sticks up an inch above the top. About<br />
how much work does it take to drink the milkshake through the<br />
straw (neglecting friction)? Answer in inch-ounces.<br />
91.3244 in.-oz<br />
y<br />
y<br />
8 y<br />
y<br />
8<br />
7<br />
y 14x 17.5<br />
20<br />
y<br />
Δy<br />
y<br />
(1.75, 7)<br />
y 17.5<br />
14<br />
(a) How much work does it take to empty the tank by pumping<br />
the water back to ground level once the tank is full? 1,497,600 ft-lb<br />
(b) If the water is pumped to ground level with a<br />
511-horsepower motor (work output 250 ft • lbsec),<br />
how long will it take to empty the full tank (to the nearest<br />
minute)? 100 min<br />
(c) Show that the pump in part (b) will lower the water level<br />
10 ft (halfway) during the first 25 min of pumping.<br />
0<br />
x<br />
1.25<br />
Dimensions in inches<br />
21. Revisiting Example 3 How much work will it take to pump<br />
the oil in Example 3 to a level 3 ft above the cone’s rim?<br />
53,482.5 ft-lb
Section 7.5 Applications from Science and Statistics 427<br />
22. Pumping Milk Suppose the conical tank in Example 3<br />
contains milk weighing 64.5 lbft 3 instead of olive oil. How<br />
26. Milk Carton A rectangular milk carton measures 3.75 in.<br />
by 3.75 in. at the base and is 7.75 in. tall. Find the force<br />
much work will it take to pump the contents to the rim? 34,582.65 ft-lb of the milk weighing 64.5 lbft 3 on one side when the<br />
23. Writing to Learn You are in charge of the evacuation and<br />
carton is full. 4.2 lb<br />
repair of the storage tank shown here. The tank is a hemisphere 27. Find the probability that a clock stopped between 1:00 and 5:00.<br />
of radius 10 ft and is full of benzene weighing 56 lbft 3 1/3<br />
.<br />
28. Find the probability that a clock stopped between 3:00 and 6:00.<br />
1/4<br />
y<br />
29. Suppose a telephone help line takes a mean of 2 minutes to<br />
answer calls. If the standard deviation is s 2, what percentage<br />
Outlet pipe<br />
of the calls are answered in the range of 0 to 4 minutes? 68%<br />
x 2 y 2 100<br />
10<br />
30. Test Scores The mean score on a national aptitude test is 498<br />
2 ft<br />
with a standard deviation of 100 points.<br />
0<br />
(a) What percentage of the population has scores between 400<br />
and 500? 0.34 (34%)<br />
10 x<br />
(b) If we sample 300 test-takers at random, about how many<br />
z<br />
should have scores above 700? 6.5<br />
A firm you contacted says it can empty the tank for 12 cent<br />
per foot-pound of work. Find the work required to empty the<br />
tank by pumping the benzene to an outlet 2 ft above the tank.<br />
If you have budgeted $5000 for the job, can you afford to hire<br />
the firm? 967,611 ft-lb, yes<br />
24. Water Tower Your town has decided to drill a well to increase<br />
its water supply. As the town engineer, you have determined that<br />
a water tower will be necessary to provide the pressure needed<br />
for distribution, and you have designed the system shown here.<br />
The water is to be pumped from a 300-ft well through a vertical<br />
4-in. pipe into the base of a cylindrical tank 20 ft in diameter<br />
and 25 ft high. The base of the tank will be 60 ft above ground.<br />
The pump is a 3-hp pump, rated at 1650 ft • lbsec. To the<br />
nearest hour, how long will it take to fill the tank the first time?<br />
(Include the time it takes to fill the pipe.) Assume water weighs<br />
62.4 lbft 3 . 31 hr<br />
10 ft<br />
25 ft<br />
31. Heights of Females The mean height of an adult female in<br />
New York City is estimated to be 63.4 inches with a standard<br />
deviation of 3.2 inches. What proportion of the adult females in<br />
New York City are<br />
(a) less than 63.4 inches tall? 0.5 (50%)<br />
(b) between 63 and 65 inches tall? 0.24 (24%)<br />
(c) taller than 6 feet? 0.0036 (0.36%)<br />
(d) exactly 5 feet tall? 0 if we assume a continuous distribution;<br />
0.071; 7.1% between 59.5 in. and 60.5 in.<br />
32. Writing to Learn Exercises 30 and 31 are subtly different, in<br />
that the heights in Exercise 31 are measured continuously and<br />
the scores in Exercise 30 are measured discretely. The discrete<br />
probabilities determine rectangles above the individual test<br />
scores, so that there actually is a nonzero probability of scoring,<br />
say, 560. The rectangles would look like the figure below, and<br />
would have total area 1.<br />
Ground<br />
4 in.<br />
Water surface<br />
NOT TO SCALE<br />
60 ft<br />
300 ft<br />
Submersible pump<br />
25. Fish Tank A rectangular freshwater fish tank with base<br />
2 4 ft and height 2 ft (interior dimensions) is filled to within<br />
2 in. of the top.<br />
(a) Find the fluid force against each end of the tank. 209.73 lb<br />
(b) Suppose the tank is sealed and stood on end (without<br />
spilling) so that one of the square ends is the base. What does<br />
that do to the fluid forces on the rectangular sides?<br />
838.93 lb; the fluid force doubles<br />
Explain why integration gives a good estimate for the<br />
probability, even in the discrete case. Integration is a good<br />
approximation to the area.<br />
33. Writing to Learn Suppose that f t is the probability density<br />
function for the lifetime of a certain type of lightbulb where t is<br />
in hours. What is the meaning of the integral<br />
800<br />
The proportion of lightbulbs<br />
that last between 100 and 800<br />
f t dt?<br />
hours.<br />
100<br />
Standardized Test Questions<br />
You may use a graphing calculator to solve the following<br />
problems.<br />
34. True or False A force is applied to compress a spring several<br />
inches. Assume the spring obeys Hooke’s Law. Twice as much<br />
work is required to compress the spring the second inch than is<br />
required to compress the spring the first inch. Justify your<br />
answer. False. Three times as much work is required.
428 Chapter 7 Applications of Definite Integrals<br />
35. True. The force against each vertical side is 842.4 lb<br />
35. True or False An aquarium contains water weighing 62.4 lb/ft 3 .<br />
The aquarium is in the shape of a cube where the length of<br />
each edge is 3 ft. Each side of the aquarium is engineered to<br />
withstand 1000 pounds of force. This should be sufficient to<br />
withstand the force from water pressure. Justify your answer.<br />
36. Multiple Choice A force of F(x) 350x newtons moves a<br />
particle along a line from x 0 m to x 5 m. Which of the following<br />
gives the best approximation of the work done by the<br />
force? E<br />
(A) 1750 J (B) 2187.5 J (C) 2916.67 J<br />
(D) 3281.25 J (E) 4375 J<br />
37. Multiple Choice A leaky bag of sand weighs 50 n. It is lifted<br />
from the ground at a constant rate, to a height of 20 m above the<br />
ground. The sand leaks at a constant rate and just finishes draining<br />
as the bag reaches the top. Which of the following gives the<br />
work done to lift the sand to the top? (Neglect the bag.) D<br />
(A) 50 J (B) 100 J (C) 250 J (D) 500 J (E) 1000 J<br />
38. Multiple Choice A spring has a natural length of 0.10 m.<br />
A 200-n force stretches the spring to a length of 0.15 m. Which<br />
of the following gives the work done in stretching the spring<br />
from 0.10 m to 0.15 m? B<br />
(A) 0.05 J (B) 5 J (C) 10 J (D) 200 J (E) 4000 J<br />
39. Multiple Choice A vertical right cylindrical tank measures<br />
12 ft high and 16 ft in diameter. It is full of water weighing<br />
62.4 lb/ft 3 . How much work does it take to pump the water to the<br />
level of the top of the tank? Round your answer to the nearest<br />
ft-lb. E<br />
(A) 149,490 ft-lb<br />
(B) 285,696 ft-lb<br />
(C) 360,240 ft-lb<br />
(D) 448,776 ft-lb<br />
(E) 903,331 ft-lb<br />
Extending the Ideas<br />
40. Putting a Satellite into Orbit The strength of Earth’s<br />
gravitational field varies with the distance r from Earth’s center,<br />
and the magnitude of the gravitational force experienced by a<br />
satellite of mass m during and after launch is<br />
Fr m MG<br />
.<br />
r<br />
2<br />
Here, M 5.975 10 24 kg is Earth’s mass,<br />
G 6.6726 10 11 N • m 2 kg 2 is the universal gravitational<br />
constant, and r is measured in meters. The work it takes to lift a<br />
1000-kg satellite from Earth’s surface to a circular orbit 35,780 km<br />
above Earth’s center is therefore given by the integral<br />
Work 35,780,000<br />
100 0MG<br />
<br />
r<br />
2<br />
dr joules.<br />
6,370,000<br />
The lower limit of integration is Earth’s radius in meters at the<br />
launch site. Evaluate the integral. (This calculation does not take<br />
into account energy spent lifting the launch vehicle or energy<br />
spent bringing the satellite to orbit velocity.) 5.1446 10 10 J<br />
41. Forcing Electrons Together Two electrons r meters apart<br />
repel each other with a force of<br />
F 23 10 29<br />
r<br />
2<br />
newton.<br />
(a) Suppose one electron is held fixed at the point 1, 0 on the<br />
x-axis (units in meters). How much work does it take<br />
to move a second electron along the x-axis from the point<br />
1, 0 to the origin? 1.15 10 28 J<br />
(b) Suppose an electron is held fixed at each of the points<br />
1, 0 and 1, 0. How much work does it take to move a third<br />
electron along the x-axis from 5, 0 to 3, 0? 7.6667 10 29 J<br />
42. Kinetic Energy If a variable force of magnitude Fx moves a<br />
body of mass m along the x-axis from x 1 to x 2 , the body’s velocity<br />
v can be written as dxdt (where t represents time). Use Newton’s<br />
second law of motion, F mdvdt, and the Chain Rule<br />
d v dv<br />
d x dv<br />
v <br />
See page 429.<br />
dt<br />
d x dt<br />
d x<br />
to show that the net work done by the force in moving the body<br />
from x 1 to x 2 is<br />
W x 2<br />
Fx dx 1 2 mv 2 2 1 2 mv 1 2 , (1)<br />
x 1<br />
where v 1 and v 2 are the body’s velocities at x 1 and x 2 . In physics<br />
the expression 12mv 2 is the kinetic energy of the body<br />
moving with velocity v. Therefore, the work done by the force<br />
equals the change in the body’s kinetic energy, and we can find<br />
the work by calculating this change.<br />
Weight vs. Mass<br />
Weight is the force that results from gravity pulling<br />
on a mass. The two are related by the equation in<br />
Newton’s second law,<br />
weight mass acceleration.<br />
Thus,<br />
newtons kilograms msec 2 ,<br />
pounds slugs ftsec 2 .<br />
To convert mass to weight, multiply by the acceleration<br />
of gravity. To convert weight to mass, divide by<br />
the acceleration of gravity.<br />
In Exercises 43–49, use Equation 1 from Exercise 42.<br />
43. Tennis A 2-oz tennis ball was served at 160 ftsec (about<br />
109 mph). How much work was done on the ball to make it<br />
go this fast? 50 ft-lb<br />
44. Baseball How many foot-pounds of work does it take to throw<br />
a baseball 90 mph? A baseball weighs 5 oz 0.3125 lb.<br />
85.1 ft-lb<br />
45. Golf A 1.6-oz golf ball is driven off the tee at a speed of<br />
280 ftsec (about 191 mph). How many foot-pounds of work<br />
are done getting the ball into the air? 122.5 ft-lb<br />
46. Tennis During the match in which Pete Sampras won the 1990<br />
U.S. Open men’s tennis championship, Sampras hit a serve that<br />
was clocked at a phenomenal 124 mph. How much work did<br />
Sampras have to do on the 2-oz ball to get it to that speed?<br />
64.6 ft-lb
Section 7.5 Applications from Science and Statistics 429<br />
47. Football A quarterback threw a 14.5-oz football 88 ftsec<br />
(60 mph). How many foot-pounds of work were done on the<br />
ball to get it to that speed? 109.7 ft-lb<br />
48. Softball How much work has to be performed on a<br />
6.5-oz softball to pitch it at 132 ftsec (90 mph)? 110.6 ft-lb<br />
42. F m d v<br />
mv d v<br />
, so W x 2<br />
F(x) dx<br />
dt<br />
dx<br />
x 1<br />
x 2<br />
mv d v<br />
dx<br />
x 1<br />
v 2<br />
mv dv 1 dx<br />
v 1<br />
2 mv 2 2 1 2 mv 1 2<br />
Quick Quiz for AP* Preparation: Sections 7.4 and 7.5<br />
49. A Ball Bearing A 2-oz steel ball bearing is placed on a vertical<br />
spring whose force constant is k 18 lbft. The spring is<br />
compressed 3 in. and released. About how high does the ball<br />
bearing go? (Hint: The kinetic (compression) energy, mgh, of a<br />
spring is 1 2 ks 2 , where s is the distance the spring is compressed,<br />
m is the mass, g is the acceleration of gravity, and h is the height.)<br />
4.5 ft<br />
You should solve the following problems without using a<br />
4. Free Response The front of a fish tank is rectangular in shape<br />
graphing calculator.<br />
and measures 2 ft wide by 1.5 ft tall. The water in the tank exerts<br />
1. Multiple Choice The length of a curve from x 0 to x 1 is pressure on the front of the tank. The pressure at any point on the<br />
given by 1 front of the tank depends only on how far below the surface the<br />
1 16<br />
0 x6 dx. If the curve contains the point (1, 4),<br />
point lies and is given by the equation p 62.4h, where h is<br />
which of the following could be an equation for this curve? A<br />
depth below the surface measured in feet and p is pressure<br />
(A) y x 4 3<br />
measured in pounds/ft 2 .<br />
(B) y x 4 1<br />
(C) y 1 16x 6<br />
(D) y 1 x 16<br />
6<br />
1.5 ft<br />
(E) y x x h<br />
7<br />
<br />
7<br />
h<br />
2. Multiple Choice Which of the following gives the length of<br />
the path described by the parametric equations x 1 4 t 4 and<br />
2 ft<br />
y t 3 , where 0 t 2? D<br />
The front of the tank can be partitioned into narrow horizontal<br />
2<br />
(A) t 6 9t 4 bands of height Δh. The force exerted by the water on a band at<br />
dt<br />
depth h<br />
0<br />
i is approximately<br />
2<br />
pressure area = 62.4h<br />
(B)<br />
i 2Δh.<br />
t 6 1 dt<br />
0<br />
(a) Write the Riemann sum that approximates the force exerted<br />
2<br />
on the entire front of the tank.<br />
(C) 1 9t 4 dt<br />
(b) Use the Riemann sum from part (a) to write and evaluate a<br />
0<br />
definite integral that gives the force exerted on the front of the<br />
2<br />
tank. Include correct units.<br />
(D) tt 6 9 4 dt<br />
0<br />
(c) Find the total force exerted on the front of the tank if the front<br />
2<br />
(and back) are semicircles with diameter 2 ft. Include correct units.<br />
(E) t 3 3t 2 dt<br />
0<br />
2 ft<br />
3. Multiple Choice The base of a solid is a circle of radius<br />
2 inches. Each cross section perpendicular to a certain diameter<br />
is a square with one side lying in the circle. The volume of the (a) n<br />
62.4h i 2 h<br />
solid in cubic inches is C<br />
i1<br />
(A) 16 (B) 16p (C) 12 8<br />
(D) 12 8p (E) 32p (b) 1.5<br />
62.4h 2 dh 140.4 lbs (c) 1.5<br />
62.4h 21h 2 dh 41.6 lbs<br />
3 3<br />
0<br />
0
430 Chapter 7 Applications of Definite Integrals<br />
Calculus at Work<br />
I am working toward becoming an<br />
archeaoastronomer and ethnoastronomer<br />
of Africa. I have a Bachelor’s degree in<br />
Physics, a Master’s degree in Astronomy,<br />
and a Ph.D. in Astronomy and Astrophysics.<br />
From 1988 to 1990 I was a member<br />
of the Peace Corps, and I taught<br />
mathematics to high school students in<br />
the Fiji Islands. Calculus is a required<br />
course in high schools there.<br />
For my Ph.D. dissertation, I investigated<br />
the possibility of the birthrate of stars<br />
being related to the composition of star<br />
formation clouds. I collected data on the<br />
absorption of electromagnetic emissions<br />
emanating from these regions. The intensity<br />
of emissions graphed versus wavelength<br />
produces a flat curve with downward<br />
spikes at the characteristic wavelengths<br />
of the elements present. An estimate<br />
of the area between a spike and the<br />
flat curve results in a concentration in<br />
molecules/cm 3 of an element. This area is<br />
the difference in the integrals of the flat<br />
and spike curves. In particular, I was looking<br />
for a large concentration of water-ice,<br />
which increases the probability of planets<br />
forming in a region.<br />
Currently, I am applying for two research<br />
grants. One will allow me to use the NASA<br />
infrared telescope on Mauna Kea to<br />
search for C 3 S 2 in comets. The other will<br />
help me study the history of astronomy in<br />
Tunisia.<br />
Jarita Holbrook<br />
Los Angeles, CA<br />
Chapter 7 Key Terms<br />
arc length (p. 413)<br />
area between curves (p. 390)<br />
Cavalieri’s theorems (p. 404)<br />
center of mass (p. 389)<br />
constant-force formula (p. 384)<br />
cylindrical shells (p. 402)<br />
displacement (p. 380)<br />
fluid force (p. 421)<br />
fluid pressure (p. 421)<br />
foot-pound (p. 384)<br />
force constant (p. 385)<br />
Gaussian curve (p. 423)<br />
Hooke’s Law (p. 385)<br />
inflation rate (p. 388)<br />
joule (p. 384)<br />
length of a curve (p. 413)<br />
mean (p. 423)<br />
moment (p. 389)<br />
net change (p. 379)<br />
newton (p. 384)<br />
normal curve (p. 423)<br />
normal pdf (p. 423)<br />
probability density function (pdf) (p. 422)<br />
68-95-99.7 rule (p. 423)<br />
smooth curve (p. 413)<br />
smooth function (p. 413)<br />
solid of revolution (p. 400)<br />
standard deviation (p. 423)<br />
surface area (p. 405)<br />
total distance traveled (p. 381)<br />
universal gravitational constant (p. 428)<br />
volume by cylindrical shells (p. 402)<br />
volume by slicing (p. 400)<br />
volume of a solid (p. 399)<br />
weight-density (p. 421)<br />
work (p. 384)<br />
Chapter 7 Review Exercises<br />
The collection of exercises marked in red could be used as a chapter<br />
test.<br />
In Exercises 1–5, the application involves the accumulation of small<br />
changes over an interval to give the net change over that entire interval.<br />
Set up an integral to model the accumulation and evaluate it to<br />
answer the question.<br />
1. A toy car slides down a ramp and coasts to a stop after<br />
5 sec. Its velocity from t 0 to t 5 is modeled by<br />
vt t 2 0.2t 3 ftsec. How far does it travel? 10.417 ft<br />
2. The fuel consumption of a diesel motor between weekly<br />
maintenance periods is modeled by the function ct <br />
4 0.001t 4 galday, 0 t 7. How many gallons does<br />
it consume in a week? 31.361 gal<br />
3. The number of billboards per mile along a 100-mile stretch of an<br />
interstate highway approaching a certain city is modeled by the<br />
function Bx 21 e 0.03x ,where x is the distance from the city<br />
in miles. About how many billboards are along that stretch of<br />
highway? 1464
Chapter 7 Review Exercises 431<br />
4. A 2-meter rod has a variable density modeled by the function<br />
rx 11 4x gm, where x is the distance in meters from the<br />
base of the rod. What is the total mass of the rod? 14 g<br />
5. The electrical power consumption (measured in kilowatts) at a<br />
factory t hours after midnight during a typical day is modeled by<br />
Et 3002 cos pt12. How many kilowatt-hours of<br />
electrical energy does the company consume in a typical day? 14,400<br />
In Exercises 6–19, find the area of the region enclosed by the lines<br />
and curves.<br />
6. y x, y 1x 2 , x 2 1<br />
7. y x 1, y 3 x 2 9 2 <br />
8. x y 1, x 0, y 0 1/6<br />
y<br />
[–2, 2] by [–1.5, 1.5]<br />
18. The Bell the region enclosed by the graphs of<br />
y 3 1x 2 and y x2 3<br />
<br />
10<br />
5.7312<br />
1<br />
√⎯x √⎯y 1<br />
0 1<br />
x<br />
[–4, 4] by [–2, 3.5]<br />
9. x 2y 2 , x 0, y 3 18<br />
10. 4x y 2 4, 4x y 16 30.375<br />
11. y sin x, y x, x p4 0.0155<br />
12. y 2 sin x, y sin 2x, 0 x p 4<br />
13. y cos x, y 4 x 2 8.9023<br />
14. y sec 2 x, y 3 x 2.1043<br />
15. The Necklace one of the smaller bead-shaped regions<br />
enclosed by the graphs of y 1 cos x and<br />
y 2 cos x 23 2p/3 1.370<br />
19. The Kissing Fish the region enclosed between the graphs of<br />
y x sin x and y x sin x over the interval p, p 4p<br />
[–5, 5] by [–3, 3]<br />
[–4, 4] by [–4, 8]<br />
16. one of the larger bead-shaped regions enclosed by the curves in<br />
Exercise 15 23 4p/3 7.653<br />
17. The Bow Tie the region enclosed by the graphs of<br />
y x 3 x<br />
x and y x 2 <br />
1<br />
(shown in the next column). 1.2956<br />
20. Find the volume of the solid generated by revolving the region<br />
bounded by the x-axis, the curve y 3x 4 , and the lines x 1<br />
and x 1 about the x-axis. 2p<br />
21. Find the volume of the solid generated by revolving the region<br />
enclosed by the parabola y 2 4x and the line y x about<br />
(a) the x-axis. 32p/3 (b) the y-axis. 128p/15<br />
(c) the line x 4. 64p/5 (d) the line y 4. 32p/3<br />
22. The section of the parabola y x 2 2 from y 0 to y 2 is<br />
revolved about the y-axis to form a bowl.<br />
(a) Find the volume of the bowl. 4p<br />
(b) Find how much the bowl is holding when it is filled to a<br />
depth of k units 0 k 2. pk 2<br />
(c) If the bowl is filled at a rate of 2 cubic units per second, how<br />
fast is the depth k increasing when k 1? 1/p
432 Chapter 7 Applications of Definite Integrals<br />
23. The profile of a football resembles the ellipse shown here (all<br />
dimensions in inches). Find the volume of the football<br />
to the nearest cubic inch. 88p 276 in 3<br />
–<br />
11<br />
— 2<br />
24. The base of a solid is the region enclosed between the graphs of<br />
y sin x and y sin x from x 0 to x p. Each cross<br />
section perpendicular to the x-axis is a semicircle with diameter<br />
connecting the two graphs. Find the volume of the solid. p 2 /4<br />
25. The region enclosed by the graphs of y e x2 , y 1, and x <br />
ln 3 is revolved about the x-axis. Find the volume of the solid<br />
generated. p(2 ln 3)<br />
26. A round hole of radius 3 feet is bored through the center of a<br />
28p/3 ft<br />
sphere of radius 2 feet. Find the volume of the piece cut out.<br />
3 29.3215 ft 3<br />
27. Find the length of the arch of the parabola y 9 x 2 that lies<br />
above the x-axis. 19.4942<br />
28. Find the perimeter of the bow-tie-shaped region enclosed<br />
between the graphs of y x 3 x and y x x 3 . 5.2454<br />
29. A particle travels at 2 units per second along the curve<br />
y x 3 3x 2 2. How long does it take to travel from<br />
the local maximum to the local minimum? 2.296 sec<br />
30. Group Activity One of the following statements is true for all<br />
k 0 and one is false. Which is which? Explain. (a) is true<br />
(a) The graphs of y k sin x and y sin kx have the same<br />
length on the interval 0, 2p.<br />
(b) The graph of y k sin x is k times as long as the graph of<br />
y sin x on the interval 0, 2p.<br />
31. Let Fx x t 4 1 dt. Find the exact length of the graph of<br />
1<br />
F from x 2 to x 5 without using a calculator. 39<br />
32. Rock Climbing A rock climber is about to haul up 100 N<br />
(about 22.5 lb) of equipment that has been hanging beneath her<br />
on 40 m of rope weighing 0.8 Nm. How much work will it take<br />
to lift<br />
(a) the equipment? 4000 J (b) the rope? 640 J<br />
(c) the rope and equipment together?<br />
0<br />
4640 J<br />
33. Hauling Water You drove an 800-gallon tank truck from the<br />
base of Mt. Washington to the summit and discovered on arrival<br />
that the tank was only half full. You had started out with a full<br />
tank of water, had climbed at a steady rate, and had taken 50<br />
minutes to accomplish the 4750-ft elevation change. Assuming<br />
that the water leaked out at a steady rate, how much work was<br />
spent in carrying the water to the summit? Water weighs 8 lbgal.<br />
(Do not count the work done getting you and the truck to<br />
the top.) 22,800,000 ft-lb<br />
y<br />
4x<br />
—–<br />
2 y 2<br />
—<br />
121 12<br />
UNITS IN INCHES<br />
11<br />
— 2<br />
1<br />
x<br />
35. No, the work going uphill is positive, but the work going downhill is<br />
negative.<br />
34. Stretching a Spring If a force of 80 N is required to hold a<br />
spring 0.3 m beyond its unstressed length, how much work does<br />
it take to stretch the spring this far? How much work does it take<br />
to stretch the spring an additional meter? 12 J, 213.3 J<br />
35. Writing to Learn It takes a lot more effort to roll a stone up<br />
a hill than to roll the stone down the hill, but the weight of the<br />
stone and the distance it covers are the same. Does this mean<br />
that the same amount of work is done? Explain.<br />
36. Emptying a Bowl A hemispherical bowl with radius 8 inches<br />
is filled with punch (weighing 0.04 pound per cubic inch) to<br />
within 2 inches of the top. How much work is done emptying<br />
the bowl if the contents are pumped just high enough to get over<br />
the rim? 113.097 in.-lb<br />
37. Fluid Force The vertical triangular plate shown below is<br />
the end plate of a feeding trough full of hog slop, weighing<br />
80 pounds per cubic foot. What is the force against the plate?<br />
426.67 lbs<br />
y<br />
–4<br />
38. Fluid Force A standard olive oil can measures 5.75 in. by<br />
3.5 in. by 10 in. Find the fluid force against the base and each<br />
side of the can when it is full. (Olive oil has a weight-density<br />
of 57 pounds per cubic foot.)<br />
base 6.6385 lb,<br />
front and back:<br />
5.7726 lb,<br />
sides 9.4835 lb<br />
OLIVE<br />
OIL<br />
39. Volume A solid lies between planes perpendicular to the<br />
x-axis at x 0 and at x 6. The cross sections between the<br />
planes are squares whose bases run from the x-axis up to the<br />
curve x y 6. Find the volume of the solid. 14.4<br />
6<br />
y<br />
2<br />
0<br />
UNITS IN FEET<br />
y x – 2<br />
4<br />
x 1/2 y 1/2 √⎯⎯6<br />
6<br />
x<br />
x
Chapter 7 Review Exercises <strong>433</strong><br />
40. Yellow Perch A researcher measures the lengths of 3-year-old<br />
yellow perch in a fish hatchery and finds that they have a mean<br />
length of 17.2 cm with a standard deviation of 3.4 cm. What<br />
proportion of 3-year-old yellow perch raised under similar<br />
conditions can be expected to reach a length of 20 cm or more?<br />
0.2051 (20.5%)<br />
41. Group Activity Using as large a sample of classmates as<br />
possible, measure the span of each person’s fully stretched hand,<br />
from the tip of the pinky finger to the tip of the thumb. Based on<br />
the mean and standard deviation of your sample, what percentage<br />
of students your age would have a finger span of more than 10<br />
inches? Answers will vary.<br />
42. The 68-95-99.7 Rule (a) Verify that for every normal pdf, the<br />
proportion of the population lying within one standard deviation<br />
of the mean is close to 68%. Hint: Since it is the same for every<br />
pdf, you can simplify the function by assuming that m 0 and<br />
s 1. Then integrate from 1 to 1. (a) 0.6827 (68.27%)<br />
(b) 0.9545 (95.45%)<br />
(b) Verify the two remaining parts of the rule.<br />
43. Writing to Learn Explain why the area under the graph of a<br />
probability density function has to equal 1. The probability that<br />
the variable has some value in the range of all possible values is 1.<br />
In Exercises 44–48, use the cylindrical shell method to find the<br />
volume of the solid generated by revolving the region bounded by<br />
the curves about the y-axis.<br />
44. y 2x, y x2, x 1 p<br />
45. y 1x, y 0, x 12, x 2 3p<br />
46. y sin x, y 0, 0 x p 2p 2<br />
47. y x 3, y x 2 3x 16p/3<br />
48. the bell-shaped region in Exercise 18 9.7717<br />
49. Bundt Cake A bundt cake (see Exploration 1, Section 7.3) has<br />
a hole of radius 2 inches and an outer radius of 6 inches at the<br />
base. It is 5 inches high, and each cross-sectional slice is<br />
parabolic.<br />
(a) Model a typical slice by finding the equation of the parabola<br />
with y-intercept 5 and x-intercepts 2. y 5 5 4 x 2<br />
(b) Revolve the parabolic region about an appropriate line<br />
to generate the bundt cake and find its volume. 335.1032 in 3<br />
50. Finding a Function Find a function f that has a continuous<br />
derivative on 0, ∞ and that has both of the following<br />
properties.<br />
i. The graph of f goes through the point 1, 1.<br />
ii. The length L of the curve from 1, 1 to any point<br />
x, f x is given by the formula L ln x f x 1.<br />
In Exercises 51 and 52, find the area of the surface generated by<br />
revolving the curve about the indicated axis.<br />
51. y tan x, 0 x p4; x-axis 3.84<br />
52. xy 1, 1 y 2; y-axis 5.02<br />
AP* Examination Preparation<br />
You may use a graphing calculator to solve the following<br />
problems.<br />
53. Let R be the region in the first quadrant enclosed by the y-axis<br />
and the graphs of y 2 sin x and y sec x.<br />
(a) Find the area of R.<br />
(b) Find the volume of the solid generated when R is revolved<br />
about the x-axis.<br />
(c) Find the volume of the solid whose base is R and whose cross<br />
sections cut by planes perpendicular to the x-axis are squares.<br />
54. The temperature outside a house during a 24-hour period is given by<br />
pt<br />
F(t) 80 10 cos <br />
1 2 <br />
,0 t 24,<br />
where F(t) is measured in degrees Fahrenheit and t is measured<br />
in hours.<br />
(a) Find the average temperature, to the nearest degree Fahrenheit,<br />
between t 6 and t 14.<br />
(b) An air conditioner cooled the house whenever the outside<br />
temperature was at or above 78 degrees Fahrenheit. For what<br />
values of t was the air conditioner cooling the house?<br />
(c) The cost of cooling the house accumulates at the rate of<br />
$0.05 per hour for each degree the outside temperature exceeds<br />
78 degrees Fahrenheit. What was the total cost, to the nearest<br />
cent, to cool the house for this 24-hour period?<br />
55. The rate at which people enter an amusement park on a given<br />
day is modeled by the function E defined by<br />
15600<br />
E(t) t 2 .<br />
24t<br />
160<br />
The rate at which people leave the same amusement park on the<br />
same day is modeled by the function L defined by<br />
9890<br />
L(t) t 2 .<br />
38t<br />
370<br />
Both E(t) and L(t) are measured in people per hour, and time t is<br />
measured in hours after midnight. These functions are valid for<br />
9 t 23, which are the hours that the park is open. At time<br />
t 9, there are no people in the park.<br />
(a) How many people have entered the park by 5:00 P.M.<br />
(t 17)? Round your answer to the nearest whole number.<br />
(b) The price of admission to the park is $15 until 5:00 P.M.<br />
(t 17). After 5:00 P.M., the price of admission to the park is $11.<br />
How many dollars are collected from admissions to the park on<br />
the given day? Round your answer to the nearest whole number.<br />
(c) Let H(t) t (E(x) L(x))dx for 9 t 23. The value of<br />
9<br />
H(17) to the nearest whole number is 3725. Find the value of<br />
H(17) and explain the meaning of H(17) and H(17) in the context<br />
of the park.<br />
(d) At what time t, for 9 t 23, does the model predict that<br />
the number of people in the park is a maximum?<br />
50. f (x) x2 2 ln x 3<br />
<br />
4